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Question Number 129070 by benjo_mathlover last updated on 12/Jan/21

$$\mathrm{Solve}2\cos \mathrm{x}\cos 2\mathrm{x}=\cos (\frac{\pi }{3}-3\mathrm{x})$$

Answered by liberty last updated on 12/Jan/21

$$\iff 2\cos \mathrm{x}\cos 2\mathrm{x}=\cos (\frac{\pi }{3}-3\mathrm{x})$$$$\iff \cos 3\mathrm{x}+\cos \mathrm{x}=\frac{1}{2}\cos 3\mathrm{x}+\frac{\sqrt{3}}{2}\sin 3\mathrm{x}$$$$\iff \cos \mathrm{x}=\frac{\sqrt{3}}{2}\sin 3\mathrm{x}-\frac{1}{2}\cos 3\mathrm{x}$$$$\iff \cos \mathrm{x}=\sin (3\mathrm{x}-\frac{\pi }{6})$$$$\iff \sin (\frac{\pi }{2}-\mathrm{x})=\sin (3\mathrm{x}-\frac{\pi }{6})$$$$\iff \mathrm{x}=\{\begin{array}{l}\frac{(3\mathrm{k}+1)\pi }{6}\\ \frac{(3\mathrm{k}+1)\pi }{3}\end{array}$$

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