Question

Решите, пожалуйста:
Cos x + sin x + sin2x + 1 = 0

Answer 1

$\cos x+\sin x+\sin 2x+1=0 \\ \cos x+\sin x+2\sin x\cos x+\cos^2x+\sin^2x=0 \\ (\cos x+ \sin x)+(\cos x +\sin x)^2=0 \\ (\cos x+\sin x)(1+\cos x+\sin x)=0$

$\cos x+\sin x=0|: \cos x \\ tg x=-1 \\ x=- \frac{\pi}{4}+\pi n,n \in Z$

$\cos x+\sin x=-1$

Формула: $a \sin x\pm b\cos x= \sqrt{a^2+b^2} \sin(x\pm \arcsin \frac{1}{\sqrt{a^2+b^2}} )$

$\sqrt{a^2+b^2}= \sqrt{2}$

$\sqrt{2}\sin (x+ \frac{\pi}{4} )=-1 \\ \sin(x+ \frac{\pi}{4} )=- \frac{1}{\sqrt{2}} \\ x+\frac{\pi}{4}=(-1)^{k+1}\cdot \frac{\pi}{4}+ \pi k,k \in Z \\ x=(-1)^{k+1}\cdot \frac{\pi}{4}-\frac{\pi}{4}+ \pi k,k \in Z$