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A variety of trigonometric equations which can be solved using inverse operations.

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Solve the following equations in degrees, for $x$ in the range $0^\\circ\\leq x\\leq720^\\circ$.

\n

Give your answers correct to the nearest degree in ascending order.

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\\[\\var{a}\\sin(x)=\\var{a-1}\\]

\n

$\\sin(x)=$ [[0]]

\n

$x=$ [[1]]$^\\circ$, [[2]]$^\\circ$, [[3]]$^\\circ$ or [[4]]$^\\circ$

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\\[\\var{b}\\cos(x)=\\var{c}\\]

\n

$\\cos(x)=$ [[0]]

\n

$x=$ [[1]]$^\\circ$, [[2]]$^\\circ$, [[3]]$^\\circ$ or [[4]]$^\\circ$

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\\[\\var{b+1}\\cos(x)=-\\var{b-1}\\]

\n

$\\cos(x)=$ [[0]]

\n

$x=$ [[1]]$^\\circ$, [[2]]$^\\circ$, [[3]]$^\\circ$ or [[4]]$^\\circ$

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\\[\\var{c+3}\\sin(x)=-\\var{c}\\]

\n

$\\sin(x)=$ [[0]]

\n

$x=$ [[1]]$^\\circ$, [[2]]$^\\circ$, [[3]]$^\\circ$ or [[4]]$^\\circ$

\n

Discard any solutions out of range.

", "extendBaseMarkingAlgorithm": true, "marks": 0, "scripts": {}, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "unitTests": []}], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Simple trig equations with degrees

"}, "advice": "

(a)

\n

\\[\\var{a}\\sin(x)=\\var{a-1}\\]

\n

Dividing both sides by $\\var{a}$ gives $\\sin(x)=\\frac{\\var{a-1}}{\\var{a}}$

\n

Hence

\n

$x=\\sin^{-1}(\\frac{\\var{a-1}}{\\var{a}})=\\var{precround((180/pi)*arcsin((a-1)/a),3)}$ is one solution.

\n

\n

We use periodicity of sin to obtain the other solutions in the range $0^\\circ\\leq x\\leq720^\\circ$.

\n

$x=180-\\var{precround((180/pi)*arcsin((a-1)/a),3)}=\\var{precround(180-(180/pi)*arcsin((a-1)/a),3)}$

\n

$x=360+\\var{precround((180/pi)*arcsin((a-1)/a),3)}=\\var{precround(360+(180/pi)*arcsin((a-1)/a),3)}$

\n

$x=540-\\var{precround((180/pi)*arcsin((a-1)/a),3)}=\\var{precround(540-(180/pi)*arcsin((a-1)/a),3)}$

\n

\n

\n

(b)

\n

\\[\\var{b}\\cos(x)=\\var{c}\\]

\n

Dividing both sides by $\\var{b}$ gives $\\cos(x)=\\frac{\\var{c}}{\\var{b}}$

\n

Hence

\n

$x=\\cos^{-1}(\\frac{\\var{c}}{\\var{b}})=\\var{precround((180/pi)*arccos((c)/b),3)}$ is one solution.

\n

\n

We use periodicity of cos to obtain the other solutions in the range $0^\\circ\\leq x\\leq720^\\circ$.

\n

$x=360-\\var{precround((180/pi)*arccos((c)/b),3)}=\\var{precround(360-(180/pi)*arccos((c)/b),3)}$

\n

$x=360+\\var{precround((180/pi)*arccos((c)/b),3)}=\\var{precround(360+(180/pi)*arccos((c)/b),3)}$

\n

$x=720-\\var{precround((180/pi)*arccos((c)/b),3)}=\\var{precround(720-(180/pi)*arccos((c)/b),3)}$

\n

\n

(c)

\n

\\[\\var{b+1}\\cos(x)=-\\var{b-1}\\]

\n

Dividing both sides by $\\var{b+1}$ gives $\\cos(x)=\\frac{-\\var{b-1}}{\\var{b+1}}$

\n

Hence

\n

$x=\\cos^{-1}(\\frac{-\\var{b-1}}{\\var{b+1}})=\\var{precround((180/pi)*arccos(-(b-1)/(b+1)),3)}$ is one solution.

\n

\n

We use periodicity of cos to obtain the other solutions in the range $0^\\circ\\leq x\\leq720^\\circ$.

\n

$x=360-\\var{precround((180/pi)*arccos(-(b-1)/(b+1)),3)}=\\var{precround(360-(180/pi)*arccos(-(b-1)/(b+1)),3)}$

\n

$x=360+\\var{precround((180/pi)*arccos(-(b-1)/(b+1)),3)}=\\var{precround(360+(180/pi)*arccos(-(b-1)/(b+1)),3)}$

\n

$x=720-\\var{precround((180/pi)*arccos(-(b-1)/(b+1)),3)}=\\var{precround(720-(180/pi)*arccos(-(b-1)/(b+1)),3)}$

\n

\n

\n

(d)

\n

\\[\\var{c+3}\\sin(x)=-\\var{c}\\]

\n

Dividing both sides by $\\var{c+3}$ gives $\\sin(x)=\\frac{-\\var{c}}{\\var{c+3}}$

\n

Hence

\n

$x=\\sin^{-1}(\\frac{-\\var{c}}{\\var{c+3}})=\\var{precround((180/pi)*arcsin(-(c)/(c+3)),3)}$.

\n

\n

Since we requre $0^\\circ\\leq x\\leq720^\\circ$, this solution is not in the given range. So we need to find the 4 required solutions as follows:

\n

\n

We use periodicity of sin to obtain the other solutions in the range $0^\\circ\\leq x\\leq720^\\circ$.

\n

$x=180-(\\var{precround((180/pi)*arcsin(-(c)/(c+3)),3)})=\\var{precround(180-(180/pi)*arcsin(-(c)/(c+3)),3)}$

\n

$x=360+(\\var{precround((180/pi)*arcsin(-(c)/(c+3)),3)})=\\var{precround(360+(180/pi)*arcsin(-(c)/(c+3)),3)}$

\n

$x=540-(\\var{precround((180/pi)*arcsin(-(c)/(c+3)),3)})=\\var{precround(540-(180/pi)*arcsin(-(c)/(c+3)),3)}$

\n

$x=720+(\\var{precround((180/pi)*arcsin(-(c)/(c+3)),3)})=\\var{precround(720+(180/pi)*arcsin(-(c)/(c+3)),3)}$

", "type": "question"}, {"name": "Simon's copy of Trigonometric Equations 2 - Simple (Radians)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Katie Lester", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/586/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}], "advice": "

(a)

\n

\\[\\var{a}\\sin(x)=\\var{a-1}\\]

\n

Dividing both sides by $\\var{a}$ gives $\\sin(x)=\\frac{\\var{a-1}}{\\var{a}}$

\n

Hence

\n

$x=\\sin^{-1}(\\frac{\\var{a-1}}{\\var{a}})=\\var{precround(arcsin((a-1)/a),3)}$ is one solution.

\n

\n

We use periodicity of sin to obtain the other solutions in the range $0\\leq\\theta\\leq2\\pi$.

\n

$x=\\pi-\\var{precround(arcsin((a-1)/a),3)}=\\var{precround(pi-arcsin((a-1)/a),3)}$

\n

\n

\n

(b)

\n

\\[\\var{b}\\cos(x)=\\var{c}\\]

\n

Dividing both sides by $\\var{b}$ gives $\\cos(x)=\\frac{\\var{c}}{\\var{b}}$

\n

Hence

\n

$x=\\cos^{-1}(\\frac{\\var{c}}{\\var{b}})=\\var{precround(arccos((c)/b),3)}$ is one solution.

\n

\n

We use periodicity of cos to obtain the other solutions in the range $0\\leq\\theta\\leq2\\pi$.

\n

$x=2\\pi-\\var{precround(arccos((c)/b),3)}=\\var{precround(2pi-arccos((c)/b),3)}$

\n

\n

\n

(c)

\n

\n

\\[\\var{c+3}\\sin(x)=-\\var{c}\\]

\n

Dividing both sides by $\\var{c+3}$ gives $\\sin(x)=\\frac{-\\var{c}}{\\var{c+3}}$

\n

Hence

\n

$x=\\sin^{-1}(\\frac{-\\var{c}}{\\var{c+3}})=\\var{precround(arcsin(-(c)/(c+3)),3)}$.

\n

\n

Since we requre $0\\leq\\theta\\leq2\\pi$, this solution is not in the given range. So we need to find the 2 required solutions as follows:

\n

\n

We use periodicity of sin to obtain the other solutions in the range $0\\leq\\theta\\leq2\\pi$.

\n

$x=\\pi-(\\var{precround(arcsin(-(c)/(c+3)),3)})=\\var{precround(pi-arcsin(-(c)/(c+3)),3)}$

\n

$x=2\\pi+(\\var{precround(arcsin(-(c)/(c+3)),3)})=\\var{precround(2pi+arcsin(-(c)/(c+3)),3)}$

\n

\n

(d)

\n

\\[\\var{b+1}\\cos(x)=-\\var{b-1}\\]

\n

Dividing both sides by $\\var{b+1}$ gives $\\cos(x)=\\frac{-\\var{b-1}}{\\var{b+1}}$

\n

Hence

\n

$x=\\cos^{-1}(\\frac{-\\var{b-1}}{\\var{b+1}})=\\var{precround(arccos(-(b-1)/(b+1)),3)}$ is one solution.

\n

\n

We use periodicity of cos to obtain the other solutions in the range $0\\leq\\theta\\leq2\\pi$.

\n

$x=2\\pi-\\var{precround(arccos(-(b-1)/(b+1)),3)}=\\var{precround(2pi-arccos(-(b-1)/(b+1)),3)}$

", "statement": "

Solve the following equations in radians, for $\\theta$ in the range $0\\leq\\theta\\leq2\\pi$.

\n

Give your answers correct to 2 decimal places in ascending order.

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Simple trig equations with radians

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\\[\\var{a}\\sin(\\theta)=\\var{a-1}\\]

\n

$\\sin(\\theta)=$ [[0]]

\n

$\\theta=$ [[1]] or [[2]]

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\\[\\var{b}\\cos(\\theta)=\\var{c}\\]

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$\\cos(\\theta)=$ [[0]]

\n

$\\theta=$ [[1]] or [[2]]

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\\[\\var{c+3}\\sin(\\theta)=-\\var{c}\\]

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$\\sin(\\theta)=$ [[0]]

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$\\theta=$ [[1]] or [[2]]

\n

Discard any solutions out of range.

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\\[\\var{b+1}\\cos(\\theta)=-\\var{b-1}\\]

\n

$\\cos(\\theta)=$ [[0]]

\n

$\\theta=$ [[1]] or [[2]]

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\\[\\var{a+1}\\cos(x-\\var{b})=\\var{a-2}\\]

\n

\n

(a)

\n

Dividing both sides by $\\var{a+1}$ gives $\\cos(x-\\var{b})=\\frac{\\var{a-2}}{\\var{a+1}}$

\n

\n

(b)

\n

Hence

\n

$x-\\var{b}=\\cos^{-1}(\\frac{\\var{a-2}}{\\var{a+1}})=\\var{precround((180/pi)*arccos((a-2)/(a+1)),3)}$ is a solution.

\n

\n

Since we want solutions $0^\\circ\\leq x\\leq720^\\circ$, this corresponds to $-\\var{b}^\\circ\\leq x-\\var{b} \\leq\\var{720-b}^\\circ$

\n

\n

We use periodicity of cos to obtain the other solutions in the range $-\\var{b}^\\circ\\leq x-\\var{b} \\leq\\var{720-b}^\\circ$:

\n

$x-\\var{b}=360-\\var{precround((180/pi)*arccos((a-2)/(a+1)),3)}=\\var{precround(360-(180/pi)*arccos((a-2)/(a+1)),3)}$

\n

$x-\\var{b}=360+\\var{precround((180/pi)*arccos((a-2)/(a+1)),3)}=\\var{precround(360+(180/pi)*arccos((a-2)/(a+1)),3)}$

\n

$x-\\var{b}=720-\\var{precround((180/pi)*arccos((a-2)/(a+1)),3)}=\\var{precround(720-(180/pi)*arccos((a-2)/(a+1)),3)}$

\n

\n

(c)

\n

$x-\\var{b}=\\var{precround((180/pi)*arccos((a-2)/(a+1)),3)} \\implies x = \\var{b} + \\var{precround((180/pi)*arccos((a-2)/(a+1)),3)}=\\var{precround(b+(180/pi)*arccos((a-2)/(a+1)),3)}$

\n

$x-\\var{b}=\\var{precround(360-(180/pi)*arccos((a-2)/(a+1)),3)}\\implies x = \\var{b} +\\var{precround(360-(180/pi)*arccos((a-2)/(a+1)),3)}=\\var{precround(b+360-(180/pi)*arccos((a-2)/(a+1)),3)}$

\n

$x-\\var{b}=\\var{precround(360+(180/pi)*arccos((a-2)/(a+1)),3)}\\implies x = \\var{b} +\\var{precround(360+(180/pi)*arccos((a-2)/(a+1)),3)}=\\var{precround(b+360+(180/pi)*arccos((a-2)/(a+1)),3)}$

\n

$x-\\var{b}=\\var{precround(720-(180/pi)*arccos((a-2)/(a+1)),3)}\\implies x = \\var{b} +\\var{precround(720-(180/pi)*arccos((a-2)/(a+1)),3)}=\\var{precround(b+720-(180/pi)*arccos((a-2)/(a+1)),3)}$

\n

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Solve the following equation in degrees, for $x$ in the range $0^\\circ\\leq x\\leq720^\\circ$.

\n

\\[\\var{a+1}\\cos(x-\\var{b})=\\var{a-2}\\]

\n

Give your final answers correct to the nearest degree in ascending order.

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Trigonometric equations with degrees

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$\\cos(x-\\var{b})=$ [[0]]

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$x-\\var{b}=$ [[0]]$^\\circ$, [[1]]$^\\circ$, [[2]]$^\\circ$ or [[3]]$^\\circ$

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$x=$ [[0]]$^\\circ$, [[1]]$^\\circ$, [[2]]$^\\circ$ or [[3]]$^\\circ$

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\\[\\var{b}\\sin(\\theta-\\var{d})=\\var{c}\\]

\n

\n

(a)

\n

Dividing both sides by $\\var{b}$ gives $\\sin(\\theta-\\var{d})=\\frac{\\var{c}}{\\var{b}}$

\n

\n

(b)

\n

Hence

\n

$\\theta-\\var{d}=\\sin^{-1}(\\frac{\\var{c}}{\\var{b}})=\\var{precround(arcsin(c/b),3)}$ is a solution.

\n

\n

Since we want solutions  $0\\leq\\theta\\leq4\\pi$, this corresponds to $-\\var{d}\\leq \\theta-\\var{d} \\leq\\var{4pi}-\\var{d}$

\n

\n

We use periodicity of sin to obtain the other solutions in the range $-\\var{d}\\leq \\theta-\\var{d} \\leq\\var{4pi}-\\var{d}$:

\n

$\\theta-\\var{d}=\\pi-\\var{precround(arcsin(c/b),3)}=\\var{precround(pi-arcsin(c/b),3)}$

\n

$\\theta-\\var{d}=2\\pi+\\var{precround(arcsin(c/b),3)}=\\var{precround(2pi+arcsin(c/b),3)}$

\n

$\\theta-\\var{d}=3\\pi-\\var{precround(arcsin(c/b),3)}=\\var{precround(3pi-arcsin(c/b),3)}$

\n

\n

(c)

\n

$\\theta-\\var{d}=\\var{precround(arcsin(c/b),3)} \\implies \\theta = \\var{precround(arcsin(c/b),3)} + \\var{d} = \\var{precround(d+arcsin(c/b),3)}$

\n

$\\theta-\\var{d}=\\var{precround(pi-arcsin(c/b),3)} \\implies \\theta =\\var{precround(pi-arcsin(c/b),3)} + \\var{d} =\\var{precround(d+pi-arcsin(c/b),3)}$

\n

$\\theta-\\var{d}=\\var{precround(2pi+arcsin(c/b),3)} \\implies \\theta =\\var{precround(2pi+arcsin(c/b),3)} + \\var{d} =\\var{precround(d+2pi+arcsin(c/b),3)}$

\n

$\\theta-\\var{d}=\\var{precround(3pi-arcsin(c/b),3)} \\implies \\theta =\\var{precround(3pi-arcsin(c/b),3)} + \\var{d} =\\var{precround(d+3pi-arcsin(c/b),3)}$

", "statement": "

Solve the following equations in radians, for $\\theta$ in the range $0\\leq\\theta\\leq4\\pi$:

\n

\\[\\var{b}\\sin(\\theta-\\var{d})=\\var{c}\\]

\n

Give your final answers correct to 2 decimal places in ascending order.

", "variable_groups": [], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Trigonometric equations with radians

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$\\sin(\\theta-\\var{d})=$ [[0]]

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$\\theta-\\var{d}=$ [[0]] , [[1]], [[2]] or [[3]]

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$\\theta=$ [[0]] , [[1]], [[2]] or [[3]]

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\\[\\var{a-1}\\sin(2x+\\var{b})=\\var{a-2}\\]

\n

\n

(a)

\n

Dividing both sides by $\\var{a-1}$ gives $\\sin(2x+\\var{b})=\\frac{\\var{a-2}}{\\var{a-1}}$

\n

\n

(b)

\n

Hence

\n

$2x+\\var{b}=\\sin^{-1}(\\frac{\\var{a-2}}{\\var{a-1}})=\\var{precround((180/pi)*arcsin((a-2)/(a-1)),3)}$ is a solution.

\n

\n

Since we want solutions $0^\\circ\\leq x\\leq360^\\circ$, this corresponds to $\\var{b}^\\circ\\leq 2x+\\var{b} \\leq\\var{720+b}^\\circ$

\n

\n

\n

We use periodicity of cos to obtain the other solutions in the range $\\var{b}^\\circ\\leq 2x+\\var{b} \\leq\\var{720+b}^\\circ$:

\n

$2x+\\var{b}=180-\\var{precround((180/pi)*arcsin((a-2)/(a-1)),3)}=\\var{precround(180-(180/pi)*arcsin((a-2)/(a-1)),3)}$

\n

$2x+\\var{b}=360+\\var{precround((180/pi)*arcsin((a-2)/(a-1)),3)}=\\var{precround(360+(180/pi)*arcsin((a-2)/(a-1)),3)}$

\n

$2x+\\var{b}=540-\\var{precround((180/pi)*arcsin((a-2)/(a-1)),3)}=\\var{precround(540-(180/pi)*arcsin((a-2)/(a-1)),3)}$

\n

\n

(c)

\n

$2x+\\var{b}=\\var{precround((180/pi)*arcsin((a-2)/(a-1)),3)} \\implies x = \\frac{\\var{precround((180/pi)*arcsin((a-2)/(a-1)),3)}-\\var{b}}{2}=\\var{precround((-b+(180/pi)*arcsin((a-2)/(a-1)))/2,3)}$

\n

$2x+\\var{b}=\\var{precround(180-(180/pi)*arcsin((a-2)/(a-1)),3)} \\implies x = \\frac{\\var{precround(180-(180/pi)*arcsin((a-2)/(a-1)),3)}-\\var{b}}{2}=\\var{precround((-b+180-(180/pi)*arcsin((a-2)/(a-1)))/2,3)}$

\n

$2x+\\var{b}=\\var{precround(360+(180/pi)*arcsin((a-2)/(a-1)),3)} \\implies x = \\frac{\\var{precround(360+(180/pi)*arcsin((a-2)/(a-1)),3)}-\\var{b}}{2}=\\var{precround((-b+360+(180/pi)*arcsin((a-2)/(a-1)))/2,3)}$

\n

$2x+\\var{b}=\\var{precround(540-(180/pi)*arcsin((a-2)/(a-1)),3)} \\implies x = \\frac{\\var{precround(540-(180/pi)*arcsin((a-2)/(a-1)),3)}-\\var{b}}{2}=\\var{precround((-b+540-(180/pi)*arcsin((a-2)/(a-1)))/2,3)}$

", "variablesTest": {"condition": "", "maxRuns": 100}, "statement": "

Solve the following equation in degrees, for $x$ in the range $0^\\circ\\leq x\\leq360^\\circ$.

\n

\\[\\var{a-1}\\sin(2x+\\var{b})=\\var{a-2}\\]

\n

Give your final answers correct to the nearest degree in ascending order.

", "metadata": {"description": "

More difficult trigonometric equations with degrees

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$\\sin(2x+\\var{b})=$ [[0]]

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$2x+\\var{b}=$ [[0]]$^\\circ$, [[1]]$^\\circ$, [[2]]$^\\circ$ or [[3]]$^\\circ$

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$x=$ [[0]]$^\\circ$, [[1]]$^\\circ$, [[2]]$^\\circ$ or [[3]]$^\\circ$

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\\[\\var{b}\\cos(3\\theta+\\var{d})=\\var{c}\\]

\n

\n

(a)

\n

Dividing both sides by $\\var{b}$ gives $\\cos(3\\theta+\\var{d})=\\frac{\\var{c}}{\\var{b}}$

\n

\n

(b)

\n

Hence

\n

$3\\theta+\\var{d}=\\cos^{-1}(\\frac{\\var{c}}{\\var{b}})=\\var{precround(arccos(c/b),3)}$ is a solution.

\n

\n

Since we want solutions   $0\\leq\\theta\\leq\\pi$, this corresponds to $\\var{d}\\leq 3\\theta+\\var{d} \\leq\\var{3pi}+\\var{d}$

\n

\n

We use periodicity of cos to obtain the other solutions in the range $\\var{d}\\leq 3\\theta+\\var{d} \\leq\\var{3pi}+\\var{d}$:

\n

$3\\theta+\\var{d}=2\\pi-\\var{precround(arccos(c/b),3)}=\\var{precround(2pi-arccos(c/b),3)}$

\n

$3\\theta+\\var{d}=2\\pi+\\var{precround(arccos(c/b),3)}=\\var{precround(2pi+arccos(c/b),3)}$

\n

\n

\n

(c)

\n

$3\\theta+\\var{d}=\\var{precround(arccos(c/b),3)} \\implies \\theta = \\frac{\\var{precround(arccos(c/b),3)} - \\var{d}}{3} = \\var{precround((-d+arccos(c/b))/3,3)}$

\n

$3\\theta+\\var{d}=\\var{precround(2pi-arccos(c/b),3)} \\implies \\theta =\\frac{\\var{precround(2pi-arccos(c/b),3)} - \\var{d}}{3} =\\var{precround((-d+2pi-arccos(c/b))/3,3)}$

\n

$3\\theta+\\var{d}=\\var{precround(2pi+arccos(c/b),3)} \\implies \\theta =\\frac{\\var{precround(2pi+arccos(c/b),3)} - \\var{d}}{3} =\\var{precround((-d+2pi+arccos(c/b))/3,3)}$

\n

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Solve the following equations in radians, for $\\theta$ in the range $0\\leq\\theta\\leq\\pi$:

\n

\\[\\var{b}\\cos(3\\theta+\\var{d})=\\var{c}\\]

\n

Give your final answers correct to 2 decimal places in ascending order.

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More difficult trigonometric equations with radians

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$\\cos(3\\theta+\\var{d})=$ [[0]]

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$3\\theta+\\var{d}=$ [[0]] , [[1]] or [[2]]

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$\\theta=$ [[0]] , [[1]] or [[2]]

\n

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Solve in radians, for $\\theta$ in the range $0\\leq \\theta\\leq2\\pi$.

\n

\n

$\\sin\\theta-\\sqrt{\\var{a}}\\cos\\theta=0$

\n

$\\tan\\theta=$ [[0]]

\n

$\\theta=$ [[1]] or [[2]]

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Solve in radians, for $\\theta$ in the range $0\\leq \\theta\\leq2\\pi$.

\n

\n

$\\var{b}\\cos^2\\theta=\\var{c}-\\var{d}\\cos\\theta$

\n

Rearrange the equation so that all terms are on the left hand side.

\n

[[0]] $=0$

\n

Use the quadratic formula or factorise the equation to get two values of $\\cos\\theta$

\n

$\\cos \\theta$=[[1]]and[[2]] (put the most positive value first)

\n

Using the above result, find four values of the $\\theta$

\n

$\\theta=$ [[3]] , [[4]],  [[5]], [[6]] (start with the smallest angle first)

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Solve in radians, for $\\theta$ in the range $0\\leq \\theta\\leq2\\pi$.

\n

\n

$\\cos^2\\theta-\\var{a}\\sin^2\\theta=1$

\n

$\\theta=$ [[0]], [[1]] or [[2]]

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Using trig identities to find solutions to equations

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Solve the following equations in radians, for $\\theta$ in the range $0\\leq \\theta\\leq2\\pi$.

\n

The square root of $a$ is written as sqrt($a$). $\\theta$ is written as theta.

\n

\n

You may want to use the following trigonometric identities.

\n

$\\tan(x)=\\frac{\\sin(x)}{\\cos(x)}$

\n

$\\sin^2(x)+\\cos^2(x)=1$

\n

$\\cos(2x)=\\cos^2(x)-\\sin^2(x)=2\\cos^2(x)-1=1-2\\sin^2(x)$

\n

$\\sin(2x)=2\\sin(x)\\cos(x)$

\n

\n

For angles, insert values in increasing order. DO NOT write your answer in an exact form with $\\pi$ but rather in decimal form to up 3 decimal places. 

\n

\n

\n

\n

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(a)

\n

Rearrange the equation to give $\\sin\\theta=\\sqrt{\\var{a}}\\cos\\theta$

\n

Divide both sides by $\\cos\\theta$ to give $\\tan\\theta=\\sqrt{\\var{a}}$

\n

Take $\\tan^{-1}\\sqrt{\\var{a}}$ to give $\\theta=\\var{precround(s11,3)}$. The second solution in the range $0\\leq \\theta\\leq2\\pi$ is given by $\\theta=\\var{precround(s11,3)}+\\pi = \\var{precround(s12,3)}$

\n

\n

\n

(b)

\n

 Rearrange the equation to give $\\var{b}\\cos^2\\theta+\\var{d}\\cos\\theta-\\var{c}=0$

\n

This is a quadratic in $\\cos\\theta$ which we can solve using the quadratic formula to give $\\cos\\theta =\\frac{-\\var{d}\\pm\\sqrt{\\var{d}^2+4\\times{\\var{b}\\times\\var{c}}}}{2\\times\\var{b}} = \\var{precround(positivex,3)}$ or $\\var{precround(negativex,3)}$

\n

Take $\\cos^{-1}\\var{precround(positivex,3)}=\\var{precround(theta1,3)}$ and $\\cos^{-1}\\var{precround(negativex,3)}=\\var{precround(theta2,3)}$.

\n

Now recall that the periodicity of cos means that if $\\theta$ is a solution then so is $\\ 2\\pi-\\theta$. Hence we obtain 4 solutions in the range $0\\leq \\theta\\leq2\\pi:$

\n

$\\theta = \\var{precround(theta1,3)}$  or  $2\\pi-\\var{precround(theta1,3)}=\\var{precround(theta3,3)}$  or  $\\var{precround(theta2,3)}$  or  $2\\pi-\\var{precround(theta2,3)}=\\var{precround(theta4,3)}$

\n

\n

\n

(c)

\n

Use the identity $\\cos^2\\theta=1-\\sin^2\\theta$ and substituting this in place of $\\cos^2\\theta$ gives $1-\\sin^2\\theta-\\var{a}\\sin^2\\theta=1$

\n

Hence $\\var{a+1}\\sin^2\\theta=0$

\n

So $\\sin^2\\theta=0$ and in turn $\\sin\\theta=0$, which has 3 solutions $\\theta = 0, \\pi, 2\\pi$ in the range $0\\leq \\theta\\leq2\\pi$

\n

\n

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Write $\\simplify{ {a} sin theta + {b}cos theta}$ in the form $\\ r \\sin(\\theta+ \\alpha)$

\n

\n

Write $r$ and $\\alpha$ correct to 3 d.p.

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Hence solve $\\simplify{{a}sin theta +{b}cos theta={c}}$. 

\n

First find the smallest value of $\\theta$ such that $\\ 0\\leq\\theta<2\\pi$.

\n

[[0]]

\n

Now find the largest value of $\\theta$ such that $\\ 0\\leq\\theta<2\\pi$.

\n

[[1]]

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(a)

\n

Compare $\\simplify{ {a}sin theta + {b}cos theta}$ with the identity $\\ r \\sin(\\theta+ \\alpha) \\equiv r \\sin \\theta \\cos \\alpha + r \\cos \\theta \\sin \\alpha$

\n

Hence $ \\var{a} = r \\cos \\alpha $

\n

and $ \\var{b} = r \\sin \\alpha $

\n

So $ r^2 = \\var{abs(a)}^2 + \\var{abs(b)}^2$

\n

$r = \\sqrt{\\var{abs(a)}^2 + \\var{abs(b)}^2} = \\var{precround(abs(r),3)}$

\n

Calculate $ \\tan^{-1} ({\\var{abs(b)}/\\var{abs(a)}}) = \\var{precround({alphatemp},3)} $

\n

So our possible values of $\\alpha$ are:

\n

$\\var{precround(alphatemp,3)}$       ,       $\\pi-\\var{precround(alphatemp,3)}=\\var{precround(pi-alphatemp,3)}$       ,       $\\pi+\\var{precround(alphatemp,3)}=\\var{precround(pi+alphatemp,3)}$       ,       $2\\pi-\\var{precround(alphatemp,3)}=\\var{precround(2pi-alphatemp,3)}$

\n

\n

Since $ r\\cos\\alpha =\\var{a} \\var{asign} 0$ and $ r \\sin \\alpha =  \\var{b} \\var{bsign} 0$ we choose $\\alpha = \\var{precround(switch(a<0 and b<0,pi+{alphatemp},b<0,2pi-{alphatemp},a<0,pi-{alphatemp},{alphatemp}),3)}$ to be in the correct quadrant.

\n

Hence $\\simplify{ {a} sin theta + {b} cos theta} \\equiv \\var{precround(r,3)} \\sin (\\simplify{theta +{precround(alpha,3)}})$

\n

\n

(b)

\n

We now have $\\var{precround(r,3)} \\sin (\\simplify{theta +{precround(alpha,3)}}) = \\var{c} $

\n

So  $ \\sin (\\simplify{theta +{precround(alpha,3)}}) = \\var{c} / \\var{precround(r,3)} = \\var{precround(c/r,3)} $

\n

We obtain 2 solutions:  $\\simplify{theta  +{precround(alpha,3)}}=  \\sin^{-1}(\\var{precround(c/r,3)}) = \\var{precround(arcsin(c/r),3)}$    or    $\\simplify{theta +{precround(alpha,3)}}=  \\pi - \\var{precround(arcsin(c/r),3)} = \\var{precround(pi - arcsin(c/r),3)}$

\n

Hence $\\ \\theta =  \\var{precround(arcsin(c/r),3)} - \\var{precround(alpha,3)} = \\var{precround(thetatemp1,3)} $     or     $\\ \\theta =  \\var{precround(pi - arcsin(c/r),3)} - \\var{precround(alpha,3)} = \\var{precround(thetatemp2,3)} $

\n

\n

Remember we require solutions such that $\\ 0\\leq\\theta<2\\pi$. Since sin is periodic with period $2 \\pi$ we can add or subtract $2 \\pi$ to our answers if necessary to obtain final answers of

\n

$\\ \\theta = \\var{precround(theta1,3)}$   or   $\\var{precround(theta2,3)}$

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