Trigonometric equation

$\cos 2x+\sin 3x=\cos 3x$

Solve the above equation and please post its solution.

Its answer is, $\frac{(4n-1)\pi}{4},\frac{(4n-1)\pi}{2},\frac{(4n+1)\pi}{4}+(-1)^{n}\sin^{-1}\left(\frac{1}{2\sqrt{2}}\right),2n\pi,n\in Z$

Thanks

#Geometry

Easy Math Editor

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Comments

$\cos 2x= \cos 3x - \sin 3x$

Square both sides. Use the double angle identity $\sin 2A = 2\sin A \cos A$ and $\cos^2 A = 1-\sin^2 A$ , we get

$1 - \sin^2 2x = 1 - \sin 6x \Leftrightarrow \sin^2 2x = \sin 6x$

Now use the triple angle identity, $\sin 3A = -4\sin^3 A + 3\sin A$ , then $\sin 6x = -4\sin^3 2x + 3\sin 2x$ .

So you get a cubic equation, $y^2 = -4y^3 + 3y$ , where $y = \sin 2x$ . Cna you take it from here?

Don't forget to check for extraneous roots (because you squared the equation from the start)!

Pi Han Goh - 4 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$