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Question Number 195042 by sonukgindia last updated on 22/Jul/23

Answered by MM42 last updated on 22/Jul/23

$$260=2^8+2^2$$$$\Rightarrow 2^{{16}^{{sin}^{2}x}}=2^8\mathrm{\&}{2}^{{16}^{{cos}^{2}x}}=2^2$$$$\Rightarrow {16}^{{sin}^{2}x}=8\mathrm{\&}{16}^{{cos}^{2}x}=2$$$$\Rightarrow {sin}^{2}x=\frac{3}{4}\mathrm{\&}{cos}^{2}x=\frac{1}{4}\Rightarrow x=k\pi \pm \frac{\pi }{3}$$$$or$$$$\Rightarrow 2^{{16}^{{sin}^{2}x}}=2^2\mathrm{\&}{2}^{{16}^{{cos}^{2}x}}=2^8$$$$\Rightarrow {16}^{{sin}^{2}x}=2\mathrm{\&}{16}^{{cos}^{2}x}=8$$$$\Rightarrow {sin}^{2}x=\frac{1}{4}\mathrm{\&}{cos}^{2}x=\frac{3}{4}\Rightarrow x=k\pi \pm \frac{\pi }{6}$$$$$$

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