Trigonometry: Solving Trigonometric Equations 3

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Name	Status	Author	Last Modified
Trigonometry: Solving Trigonometric Equations 3	Ready to use	Ben McGovern	10/10/2022 13:12
Trigonometry: Solving Trigonometric Equations 2	Ready to use	Ben McGovern	10/10/2022 13:12
Trigonometry: Solving Trigonometric Equations 1a	Ready to use	Ben McGovern	10/10/2022 13:12
Trigonometry: Solving Trigonometric Equations 1b	Ready to use	Ben McGovern	07/09/2023 08:56

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Question statement

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

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1. Gap Mathematical expression
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$x=$ Unnamed gap (Give your answer to 3 decimal places, or in terms of $\pi$ .)

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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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Trigonometry: Solving Trigonometric Equations 3

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Christian Lawson-Perfect 2 years, 7 months ago

Ben McGovern 3 years, 8 months ago

Ben McGovern 3 years, 8 months ago

Ben McGovern 4 years, 7 months ago

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