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

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

Give your final answers correct to the nearest degree.

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

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Solve the equation:

\\[\\var{a}\\sin(2y-\\var{n})=\\var{b}\\]

for values of $y$ between $0$ and $2\\pi$.

Write your answers, in radians, correct to 1 decimal place.

$y=$ [[0]], [[1]], [[2]], [[3]]

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Consider the equation

\\[\\simplify{{a}*(sin(theta))^2+{b}*sin(theta)}=\\var{-c}\\]

where $\\theta$ is in radians, in the range $0\\leq\\theta<2\\pi$.

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Move all terms to left side:

\\[\\simplify{{a}*(sin(theta))^2+{b}*sin(theta)+{c}}=0\\]

Using the quadratic formula (or otherwise):

\\[\\sin(\\theta)=\\frac{-(\\var{b})\\pm\\sqrt{(\\var{b})^2-4(\\var{a})(\\var{c})}}{2(\\var{a})}=\\frac{\\var{-b}\\pm\\sqrt{\\var{det}}}{\\var{2*a}}=\\var[fractionNumbers]{st1}\\text{ or }\\var[fractionNumbers]{st2}\\]

Now use inverse sine:

\\[\\theta=\\sin^{-1}\\left(\\var[fractionNumbers]{st1}\\right)=\\var{precround(t0,2)}\\text{ or }\\var{precround(t1,2)}\\text{ or }\\var{precround(t2,2)}\\]

\\[\\theta=\\sin^{-1}\\left(\\var[fractionNumbers]{st2}\\right)=\\var{precround(t3,2)}\\text{ or }\\var{precround(t4,2)}\\text{ or }\\var{precround(t5,2)}\\]

Finally discard any values which are lower than $0$ or higher than $2\\pi$.

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Enter two possible values of $\\sin(\\theta)$, correct to at least 3 decimal places. You may enter fractions.

$\\sin(\\theta)=\\;$[[0]] or [[1]]

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Hence solve the equation for $\\theta$. Include all solutions in the range $0\\leq\\theta<2\\pi$ and round each angle to 2 decimal places.

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

Poor rounding or inaccurate answer.

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Poor rounding or inaccurate answer.

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Poor rounding or inaccurate answer.

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Poor rounding or inaccurate answer.

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Solve the equation, in degrees, for $x$ in the range $0^\\circ\\leq x<360^\\circ$.

\\[\\simplify{{a}*(sin(x))^2}=\\simplify{{b}*cos(x)+{c+a}}\\]

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By using an appropriate trigonometric identity, select the equivalent equation below.

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Which of the following is equivalent to $\\sin^2(x)$?

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Now rewrite $\\sin^2(x)$ accordingly and bring all terms to one side.

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Enter two possible values of $\\cos(x)$. Do not round. You may enter your answer as a fraction.

$\\cos(x)=\\;$[[0]] or [[1]]

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In the quadratic equation in part a), what is the coefficient $a$?

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In the quadratic equation in part a), what is the coefficient $b$?

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In the quadratic equation in part a), what is the coefficient $c$?

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Now use the formula to solve the quadratic equation:

\\[\\cos(x)=\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}\\]

Hence solve the equation for $x$. Include all solutions in the range $0^\\circ\\leq x<360^\\circ$ and round all angles to the nearest degree.

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

Which method can be used when calculating inverse cosine in degrees?

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Calculate $\\simplify[fractionNumbers]{arccos({st1})}$ and discard any answers outside the range $0^\\circ\\leq x<360^\\circ$.

Then calculate $\\simplify[fractionNumbers]{arccos({st2})}$ and discard any answers outside the range $0^\\circ\\leq x<360^\\circ$.

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

\\[\\simplify{{2*A}*sin(theta)*cos(theta)}=\\simplify{{B}*sin(theta)}\\]

Round each value to the nearest whole degree.

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\\[\\var{A}\\sin(2\\theta)=\\simplify{{B}*sin(theta)}\\]

Use the identity $\\sin(2\\theta)=2\\sin(\\theta)\\cos(\\theta)$:

\\[\\var{2*A}\\sin(\\theta)\\cos(\\theta)=\\simplify{{B}*sin(theta)}\\]

Bring all terms to one side:

\\[\\simplify{{2*A}*sin(theta)*cos(theta)-{B}*sin(theta)}=0\\]

Take out the common factor of $\\sin(\\theta)$:

\\[\\sin(\\theta)\\left(\\simplify{{2*A}*cos(theta)-{B}}\\right)=0\\]

Therefore

\\[\\sin(\\theta)=0\\quad\\text{or}\\quad\\simplify{{2*A}*cos(theta)-{B}}=0\\]

The first equation gives $\\theta=\\sin^{-1}(0)=0^\\circ, 180^\\circ, 360^\\circ$.

Solve the second equation:

\\[\\cos(\\theta)=\\var[fractionNumbers]{B/(2*A)}\\]

Therefore $\\theta=\\cos^{-1}(\\var[fractionNumbers]{B/(2*A)})=\\var{precround(p1,0)}^\\circ, \\var{precround(p2,0)}^\\circ$.

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$\\theta=\\;$[[0]]$^\\circ,\\;$[[1]]$^\\circ,\\;$[[2]]$^\\circ,\\;$[[3]]$^\\circ,\\;$[[4]]$^\\circ$

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The voltage in an AC circuit is given by the formula

\\[V=\\var{R}\\sin{(\\simplify{{n}*pi*t+{a}})}\\]

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Identify the amplitude of the wave.

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Calculate the period of the wave, in seconds.

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Calculate the frequency of the wave, in Hertz.

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Determine the phase angle, in degrees.

[[0]] by [[1]]$^\\circ$

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Wrong sign. A negative phase angle is accounted for when you select \"lagging\".

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You were asked to give your answer in degrees.

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Calculate the voltage after $\\var{t}$ milliseconds.

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Calculate the first time (to the nearest millisecond) when the voltage equals $\\var{Vf}$ V.

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Incorrect, you need to enter the first positive time.

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Convert the time into milliseconds.

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Enter the derivative of the function.

$\\frac{dV}{dt}=\\;$[[0]]

$\\pi$ is part of the coefficient of $t$ so this needs to be multiplied too.

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Calculate the rate of change of voltage after $\\var{t4}$ milliseconds, and select the appropriate unit.

[[0]] [[1]]

Have you converted into seconds?

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The graph below shows the current $I$ in an alternating circuit against the time in milliseconds.

{diagram}

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Identify the following:

Amplitude: [[1]] amps

Time period: [[0]] seconds

Frequency: [[2]] Hz

Angular frequency: [[3]] rad/s

Give your answer in seconds.

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Hence write down a formula for $I$ in the form $A\\sin(\\omega t)$.

$I=$ [[0]]

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Calculate the time, in milliseconds, when the current first reaches $\\var{amplitude/2}$ amps.

for this part, give your answer in milliseconds.

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