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# 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

Note by Akshat Sharda
7 months, 1 week ago

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

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)! · 7 months, 1 week ago