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

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

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$\\tan\\theta=$ [[0]]

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$\\theta=$ [[1]] or [[2]]

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

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

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

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[[0]] $=0$

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

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

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

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

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

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

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$\\theta=$ [[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 $0\\leq \\theta\\leq2\\pi$.

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

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You may want to use the following trigonometric identities.

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

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

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

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

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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. 

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(a)

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

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

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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 $\\theta=\\var{precround(s11,3)}+\\pi = \\var{precround(s12,3)}$

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(b)

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

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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 $\\var{precround(negativex,3)}$

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

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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 $0\\leq \\theta\\leq2\\pi:$

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

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(c)

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

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Hence $\\var{a+1}\\sin^2\\theta=0$

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

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