Question

Решите уравнения по теме однородные ! 1) sin^2x=1/4; 2) 2sin^2x+sinx-1=0

Answer 1

$sin^2(x)=\frac{1}{4}\\\\ \frac{1-cos(2x)}{2}=\frac{1}{4}\\\\ 2(1-cos(2x))=1\\\\ 2-cos(2x)=1\\\\ -cos(2x)=-1\\\\ cos(2x)=1\\\\ 2x=2\pi n,\ n\in Z\\\\ x=\pi n,\ n\in Z$

$2sin^2(x)+sin(x)-1=0\\\\ 2sin^2(x)+2sin(x)-sin(x)-1=0\\\\ 2sin(x)*[sin(x)+1]-[sin(x)+1]=0\\\\ \ [2sin(x)-1]*[sin(x)+1]=0\\\\ 2sin(x)-1=0\ or\ sin(x)+1=0\\\\ sin(x)=\frac{1}{2}\ or\ sin(x)=-1\\\\ x=(-1)^k*\frac{\pi}{6}+\pi k,\ k\in Z\ \ \ or\ \ \ x=-\frac{\pi}{2}+2\pi n,\ n\in Z$

ответ: $(-1)^k*\frac{\pi}{6}+\pi k,\ k\in Z\ ;\ -\frac{\pi}{2}+2\pi n,\ n\in Z$