Question

1+sinx+cosx+sin2x+cos2x=0

Answer 1

$1+\sin \leftx\right+\cos \leftx\right+\sin \left2x\right+\cos \left2x\right=0 \\ 1+\cos \left2x\right+\cos \leftx\right+\sin \leftx\right+2\cos x\leftx\right\sin x\leftx\right=0 \\ \cos \left x\right +\sin \left x\right+2\cos ^2\leftx\right+2\cos \left x\right\sinx \leftx\right=0 \\ \left(2\cos \left(x\right)+1\right)\left(\sin \left(x\right)+\cos \left(x\right)\right)=0$


2cos(x)+1=0             sin(x)+cos(x)=0
cos(x)= - 1/2             

$\cos \left(x\right)=-\frac{1}{2} ;x=\frac{2\pi }{3}+2\pi n,\:\quad x=\frac{4\pi }{3}+2\pi n \\ sin(x)+cos(x)=0;\quad x=\frac{3\pi }{4}+\pi n$

$Otvet:x=\frac{2\pi }{3}+2\pi n,\:x=\frac{3\pi }{4}+\pi n,\:x=\frac{4\pi }{3}+2\pi n$