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Question Number 86946 by A8;15: last updated on 01/Apr/20

Commented by john santu last updated on 01/Apr/20

$${\mathrm{A}}^2\sin {}^2\mathrm{Ax}={\mathrm{B}}^2\sin \mathrm{Bx}$$$$\Rightarrow \sin {}^2\mathrm{Ax}=\frac{{\mathrm{B}}^2}{{\mathrm{A}}^2}\sin {}^2\mathrm{Bx}$$$$\cos {}^2\mathrm{Ax}=\cos {}^2\mathrm{Bx}$$$$\Rightarrow 1-\sin {}^2\mathrm{Ax}=1-\sin {}^2\mathrm{Bx}$$$$\Rightarrow \frac{{\mathrm{B}}^2}{{\mathrm{A}}^2}\sin {}^2\mathrm{Bx}=\sin {}^2\mathrm{Bx}$$$$\Rightarrow \mathrm{B}=\mathrm{A}\vee \mathrm{B}=-\mathrm{A}$$$$$$

Commented by john santu last updated on 01/Apr/20

$$\mathrm{A}=\mathrm{B}\Rightarrow 2\cos \mathrm{Ax}=0$$$$\cos \mathrm{Ax}=0\Rightarrow \mathrm{Ax}=\pm \frac{\pi }{2}+2\mathrm{k}\pi$$$$\mathrm{x}=\pm \frac{\pi }{2\mathrm{A}}+\frac{2\mathrm{k}\pi }{\mathrm{A}}$$

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