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Question Number 134331 by Mikael_786 last updated on 02/Mar/21

$${\cos }^2(\frac{3\pi }{2}-\mathrm{x})={\cos }^2(\mathrm{x}+\pi )$$$$0⩽\mathrm{x}⩽\pi$$$$\mathrm{x}=?$$

Answered by EDWIN88 last updated on 02/Mar/21

$$\cos (\frac{3\pi }{2}-\mathrm{x})=-\sin \mathrm{x};\cos (\mathrm{x}+\pi )=-\cos \mathrm{x}$$$$\Rightarrow {(-\sin \mathrm{x})}^2={(-\cos \mathrm{x})}^2$$$$\Rightarrow (\sin \mathrm{x}+\cos \mathrm{x})(\sin \mathrm{x}-\cos \mathrm{x})=0$$$$\{\begin{array}{l}\sin \mathrm{x}-\cos \mathrm{x}=0\Rightarrow \mathrm{x}=\frac{\pi }{4}\\ \sin \mathrm{x}=-\cos \mathrm{x}\Rightarrow \mathrm{x}=\frac{3\pi }{4}\end{array}$$

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