Помогите пожалуйста решить ) 1)(cos + sinx)²=1+sinx * cosx 2)cos x/2 +1 =0 3)cosx +...

Решите задачу:

$sin^2x+cos^2x=1\\\\1) \; (cosx+ sinx)^2=1+sinxcosx\\cos^2x+2sinxcosx+sin^2x=1+sinxcosx\\(sin^2x+cos^2x)+2sinxcosx-1-sinxcosx=0\\1-1+sinxcosx=0\\sinxcosx=0\\\\sinx=0\\x=\pi n, \; n\in Z;\\\\cosx=0\\x=\frac{\pi}{2}+\pi k, \; k\in Z.$

$2)\; cos\frac{x}{2} +1 =0\\cos\frac{x}{2}=-1\\\frac{x}{2}=\pi+2\pi n\\x=2\pi + 4\pi n, \; n\in Z;\\\\3)\; cosx + sin(\frac{\pi}{2} - x) + cos (\pi +x)=0\\cosx+cosx-cosx=0\\cosx=0\\x=\frac{\pi}{2}+\pi n, \; n\in Z;$

$4)\;4cos^2x-3 =0\\cos^2x=\frac{3}{4}\\cosx=\pm\frac{\sqrt3}{2}\\\\cosx=\frac{\sqrt3}{2}\\x=\pm arccos\frac{\sqrt3}{2}+2\pi n\\x=\pm \frac{\pi}{6}+2\pi n, \; n\in Z;\\\\cosx=-\frac{\sqrt3}{2}\\x=\pm(\pi-arccos\frac{\sqrt3}{2})=2\pi n\\x=\pm(\pi-\frac{\pi}{6})+2\pi n\\x=\pm\frac{5\pi}{6}+2\pi n, \; n\in Z.$