Question

решите уравнение cos3x+cosx+2sin^2x=1

Answer 1

$cos3x+cosx+2sin^2x=1\\\\2\cdot cos\frac{3x+x}{2}\cdot cos\frac{3x-x}{2}+2sin^2x=sin^2x+cos^2x\\\\2\cdot cos2x\cdot cosx+\underbrace {sin^2x-cos^2x}_{-(cos^2x-sin^2x)}=0\; \; ,\; \; cos^2x-sin^2x=cos2x\\\\2\cdot cos2x\cdot cosx-cos2x=0\\\\cos2x\cdot (2cosx-1)=0\\\\a)\; \; cos2x=0\; ,\; \; 2x=\frac{\pi}{2}+\pi n\; \; ,\; \; x=\frac{\pi}{4}+\frac{\pi n}{2}\; ,\; n\in Z\\\\b)\; \; cosx=\frac{1}{2}\; ,\; \; x=\pm \frac{\pi }{3}+2\pi k\; ,\; k\in Z\\\\Otvet:\; \; x=\frac{\pi }{4}+\frac{\pi n}{2}\; ,\; x=\pm \frac{\pi }{3}+2\pi k\; ,\; \; n,k\in Z$