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Question Number 167486 by greogoury55 last updated on 18/Mar/22

$$\cos {}^{2}x=\cos \left(\frac{4x}{3}\right);0⩽x⩽2\pi$$

Answered by mr W last updated on 18/Mar/22

$$\cos 2x+1=2\cos \frac{4x}{3}$$$$\cos \frac{6x}{3}+1=2\cos \frac{4x}{3}$$$$lett=\frac{2x}{3}$$$$\cos 3t+1=2\cos 2t$$$$4{\cos }^{3}t-3\cos t+1=4{\cos }^{2}t-2$$$$(4{\cos }^{2}t-3)(\cos t-1)=0$$$$\Rightarrow \cos t=1\Rightarrow t=2k\pi$$$$\Rightarrow \cos t=\pm \frac{\sqrt{3}}{2}\Rightarrow t=k\pi \pm \frac{\pi }{6}$$$$\Rightarrow x=3k\pi$$$$\Rightarrow x=\frac{3k\pi }{2}\pm \frac{\pi }{4}$$

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