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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
2 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)! Pi Han Goh · 2 months, 1 week ago

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