[tex] \frac{ 9^{sin2x}-3^{2 \sqrt{2} sinx } }{ \sqrt{11sinx} } =0[/tex]. Created by elvira1234123. algebra-ru

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August 2022 1 16 Report

[tex] \frac{ 9^{sin2x}-3^{2 \sqrt{2} sinx } }{ \sqrt{11sinx} } =0[/tex]

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$\frac{9^{sin2x}-3^{2\sqrt2sinx}}{\sqrt{11sinx}}=0\; ,\\\\ODZ:\; \; sinx \ \textgreater \ 0\; ,\; \; x\in (2\pi m\; ;\; \pi +2\pi m)\; ,\; m\in Z\\\\(3^2)^{sin2x}-3^{2\sqrt2sinx}=0\\\\3^{2sin2x}=3^{2\sqrt2sinx}\\\\2sin2x=2\sqrt2sinx\\\\4sinx\cdot cosx-2\sqrt2sinx=0\\\\2\sqrt2sinx\cdot (\sqrt2cosx-1)=0\\\\a)\; \; sinx=0\; \; \to \; \; \; x=\pi n,\; n\in Z\\\\b)\; \; cosx=\frac{1}{\sqrt2}=\frac{\sqrt2}{2}\\\\x=\pm \frac{\pi}{4}+2\pi k,\; k\in Z$

$c)\; \; \left \{ {{x=\pi n\; \; ili\; \; x=\pm \frac{\pi }{4}+2\pi k} \atop {x\in (2\pi m\; ;\; \pi +2\pi m)}} \right. \; \; \; \Rightarrow x=\frac{\pi}{4}+2\pi k,\; k\in Z$

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