tex]...

tornado1357 @tornado1357

July 2022 1 6 Report

[tex] \sqrt{2}sin^2( \pi/2 +x)=sin(3 \pi /2+x)[/tex]

Answers & Comments

konrad509 $\sqrt2\sin^2\left(\dfrac{\pi}{2}+x\right)=\sin\left(\dfrac{3\pi}{2}+x\right)\\
\sqrt2 \cos^2 x=-\cos x\\\
\sqrt2\cos^2 x+\cos x=0\\
\cos x(\sqrt 2\cos x+1)=0\\\\
\cos x=0\\
x=\dfrac{\pi}{2}+k\pi\\\\
\sqrt 2\cos x+1=0\\
\sqrt 2 \cos x=-1\\
\cos x=-\dfrac{1}{\sqrt2}\\
\cos x=-\dfrac{\sqrt2}{2}\\
x=-\dfrac{3\pi}{4}+2k\pi \vee x=\dfrac{3\pi}{4}+2k\pi \\\\
\boxed{x=\left\{\dfrac{\pi}{2}+k\pi,-\dfrac{3\pi}{4}+2k\pi,x=\dfrac{3\pi}{4}+2k\pi,(k\in\mathbb{Z})\right\}}
$

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