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Solve the following equations in radians, for $\\theta$ in the range $0\\leq \\theta\\leq2\\pi$.
\nThe square root of $a$ is entered as \"sqrt($a$)\". $\\theta$ is entered as \"theta\".
\n\nYou 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\nFor angles, insert values in increasing order. DO NOT write your answer in an exact form with $\\pi$ but rather in decimal form up to 3 decimal places.
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\nDivide both sides by $\\cos\\theta$ to give $\\tan\\theta=\\sqrt{\\var{a}}$
\nTake $\\tan^{-1}\\sqrt{\\var{a}}$ 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\nb) Rearrange the equation to give $\\var{b}\\cos^2\\theta+\\var{d}\\cos\\theta-\\var{c}=0$
\nThis 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 $\\var{precround(negativex,3)}$
\nTake $\\cos^{-1}\\var{precround(positivex,3)}=\\var{precround(theta1,3)}$ and $\\cos^{-1}\\var{precround(negativex,3)}=\\var{precround(theta2,3)}$.
\nNow 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\nc) Use the identity $\\cos^2\\theta=1-\\sin^2\\theta$ and substituting this in place of $\\cos^2\\theta$ gives $1-\\sin^2\\theta-\\var{a}\\sin^2\\theta=1$
\nHence $\\var{a+1}\\sin^2\\theta=0$
\nSo $\\sin^2\\theta=0$ 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", "metadata": {"description": "Using trig identities to find solutions to equations
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\n$\\tan\\theta=$ [[0]]
\n$\\theta=$ [[1]] or [[2]]
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\nRearrange the equation so that all terms are on the left hand side.
\n[[0]] $=0$
\nUse 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)
\nUsing 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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