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Question Number 131661 by benjo_mathlover last updated on 07/Feb/21

$$\log {}_{\sqrt{2}\sin \mathrm{x}}(1+\cos \mathrm{x})=2$$$$-\frac{2\pi }{3}⩽\mathrm{x}⩽\frac{\pi }{3}$$

Answered by liberty last updated on 07/Feb/21

$$\{\begin{array}{l}\sqrt{2}\sin \mathrm{x}>0;\mathrm{x}\mathrm{in}\mathrm{I}\mathrm{or}\mathrm{II}\mathrm{quadrant}\\ \sqrt{2}\sin \mathrm{x}\ne 1\Rightarrow \sin \mathrm{x}\ne \frac{1}{\sqrt{2}}\end{array}$$$$\iff 1+\cos \mathrm{x}={\left(\sqrt{2}\sin \mathrm{x}\right)}^2$$$$\iff 2\sin {}^2\mathrm{x}-\cos \mathrm{x}-1=0$$$$\iff 2-2\cos {}^2\mathrm{x}-\cos \mathrm{x}-1=0$$$$\iff 2\cos {}^2\mathrm{x}+\cos \mathrm{x}-1=0$$$$\iff (2\cos \mathrm{x}-1)(\cos \mathrm{x}+1)=0$$$$\{\begin{array}{l}\cos \mathrm{x}=\frac{1}{2}\Rightarrow \mathrm{x}=\frac{\pi }{3}\leftarrow \mathrm{only}\mathrm{solution}\\ \cos \mathrm{x}=-1\left(\mathrm{reject}\right)\end{array}$$

Commented by liberty last updated on 07/Feb/21

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