Решить уравнение: sin(x)+cos(x)=1+sin(2x)

sinx+cosx=1+2sin(2x)

sinx+cosx=1+2sinxcosx

sinx+cosx=(sinx)^2+(cosx)^2+2sinxcosx

sinx+cosx=(sinx+cosx)^2

1) sinx+cosx=1

$\sqrt{2}(\frac{\sqrt{2}}{2}sinx+\frac{\sqrt{2}}{2}cosx)=1$

$\sqrt{2}(cos\frac{\pi}{4}sinx+sin\frac{\pi}{4}cosx)=1$

$\sqrt{2}sin(x+\frac{\pi}{4})=1$

$sin(x+\frac{\pi}{4})=\frac{\sqrt{2}}{2}$

$x+\frac{\pi}{4}=\frac{\pi}{4}+2n\pi$

$x=(-1)^{n}*(-\frac{\pi}{4})+n\pi$

2) sinx+cosx=0

sinx=-cosx

sinx/cosx=-1

tgx=-1

x=-Pi/4+Pi*n (n Є N)