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Solving $\\sin(3x)=\\sin(x)$ for $x\\in \\left(0,\\frac{\\pi}{2}\\right)$.

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Solve $\\sin(3x)=\\sin(x)$, for $x \\in \\left(0,\\frac{\\pi}{2}\\right)$.

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To solve \\[ \\sin(3x)=\\sin(x), \\]

we first want to rewrite the the left-hand side using the following trigonometric identities:

$\\sin(A+B) = \\sin(A)\\cos(B) + \\sin(B)\\cos(A)$;
$\\sin(2x) = 2\\sin(x)\\cos(x)$;
$\\cos(2x) = \\cos^2(x) - \\sin^2(x)$.

\\[ \\begin{split} \\sin(3x) &\\,= \\sin(2x)\\cos(x) + \\sin(x)\\cos(2x), \\\\ &\\,= \\left(2\\sin(x)\\cos(x)\\right)\\cos(x) + \\sin(x) \\left(\\cos^2(x)-\\sin^2(x)\\right) \\\\ &\\,= 2\\sin(x) \\cos^2(x) + \\sin(x) \\cos^2(x) - \\sin^3(x) \\\\ &\\,= 3 \\sin(x) \\cos^2(x) - \\sin^3(x). \\end{split} \\]

Hence,

\\[ \\begin{split} 3 \\sin(x) \\cos^2(x) - \\sin^3(x) &\\,= \\sin(x) \\\\ 3\\cos^2(x) - \\sin^2(x) &\\,= 1. \\end{split} \\]

To make the left-hand side only in terms of $\\cos(x)$, we can use the identity \\[ \\sin^2(x) + \\cos^2(x) = 1 \\implies \\sin^2(x) = 1 - \\cos^2(x).\\]

Therefore,

\\[ \\begin{split} 3\\cos^2(x) - \\sin^2(x) &\\,= 1 \\\\ 3\\cos^2(x) - \\left(1-\\cos^2(x)\\right) &\\,=1 \\\\ 4\\cos^2(x) - 1 &\\,=1. \\end{split} \\]

Rearranging,

\\[ \\begin{split} \\cos^2(x) &\\,= \\frac{1}{2} \\\\ \\cos(x) &\\,= \\frac{1}{\\sqrt{2}} \\\\ x &\\, = \\cos^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right). \\end{split} \\]

So our solution is \\[ \\begin{split} x &\\, = \\cos^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right) \\\\ &\\,= \\frac{\\pi}{4} \\\\ &\\,= 0.785 \\, (3\\text{ d.p.}). \\end{split} \\]

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$x=$[[0]] (Give your answer to 3 decimal places, or in terms of $\\pi$.)

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