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Question Number 74945 by chess1 last updated on 04/Dec/19

Commented by chess1 last updated on 04/Dec/19

$$\mathrm{solve}\mathrm{equation}$$

Answered by MJS last updated on 04/Dec/19

$$\mathrm{solving}\mathrm{for}0⩽x<2\pi \mathrm{at}\mathrm{first}$$$$\cos x>0\wedge \sin x>0\Rightarrow 0<x<\frac{\pi }{2}$$$$\frac{\ln 24\cos x}{\ln 24\sin x}=\frac{3}{2}$$$$\Rightarrow$$$${\cos }^{2}x-{24\sin }^{3}x=0$$$$\Rightarrow$$$${\sin }^{3}x+\frac{1}{24}{\sin }^{2}x-\frac{1}{24}=0$$$$(\sin x-\frac{1}{3})({\sin }^{2}x+\frac{3}{8}\sin x+\frac{1}{8})=0$$$$\Rightarrow \sin x=\frac{1}{3}\Rightarrow x=\arcsin \frac{1}{3}$$$$\mathrm{generally}x=2n\pi +\arcsin \frac{1}{3}$$

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