Question

sin(2x)+1=cos(x)+2sin(x)



tan*( 5x - 2) = -корень 3

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Answer 1

$1)\; \; sin2x+1=cosx+2\, sinx\\\\2\, sinx\cdot cosx-2\, sinx-cosx+1=0\\\\2\, sinx(cosx-1)-(cosx-1)=0\\\\(cosx-1)(2\, sinx-1)=0\\\\a)\; \;cosx-1=0\; ,\; \; cosx=1\; ,\; \; x=2\pi n,\; n\in Z\\\\b)\; \; 2\, sinx-1=0\; ,\; \; sinx=\frac{1}{2}\; ,\; \; x=(-1)^{k}\cdot \frac{\pi}{6}+\pi k\; ,\; k\in Z\\\\Otvet:\; \; x=2\pi n\; ,\; \; x=(-1)^{k}\cdot \frac{\pi}{6}+\pi k\; ,\; n,k\in Z\; .$

$2)\; \; tg(5x-2)=-\sqrt3\\\\5x-2=arctg(-\sqrt3)+\pi n\; ,\; n\in Z\\\\5x-2=-\frac{\pi}{3}+\pi n\; ,\; n\in Z\\\\5x=2-\frac{\pi}{3}+\pi n\; ,\; n\in Z\\\\x=\frac{2}{5}-\frac{\pi}{15}+\frac{\pi n}{5}\; ,\; n\in Z$