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Question Number 130819 by EDWIN88 last updated on 29/Jan/21

$$8\sin {}^3(x+\frac{\pi }{6})=\cos \left(3x\right)$$$$x=?$$

Answered by mr W last updated on 29/Jan/21

$$letu=x+\frac{\pi }{6}$$$$\Rightarrow 3x=3u-\frac{\pi }{2}$$$$\cos \left(3x\right)=\cos (3u-\frac{\pi }{2})=\sin \left(3u\right)$$$$=3\sin u-4{\sin }^{3}u$$$$8{\sin }^{3}u=3\sin u-4{\sin }^{3}u$$$$(4{\sin }^{2}u-1)\sin u=0$$$$\Rightarrow \sin u=0\Rightarrow u=k\pi \Rightarrow x=k\pi -\frac{\pi }{6}$$$$$$$$\Rightarrow 4{\sin }^{2}u-1=0\Rightarrow \sin u=\pm \frac{1}{2}$$$$\Rightarrow u=2k\pi +\frac{\pi }{2}\pm \frac{\pi }{3}\Rightarrow x=2k\pi +\frac{\pi }{3}\pm \frac{\pi }{3}$$$$\Rightarrow u=2k\pi -\frac{\pi }{2}\pm \frac{\pi }{3}\Rightarrow x=2k\pi -\frac{2\pi }{3}\pm \frac{\pi }{3}$$$$$$$$summary:$$$$x=k\pi -\frac{\pi }{6}$$$$x=k\pi$$$$x=2k\pi -\frac{\pi }{3}$$$$x=2k\pi +\frac{2\pi }{3}$$

Commented by EDWIN88 last updated on 29/Jan/21

$$nice$$

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