Синус + косинус = 1(написать решение)

$sin(x)+cos(x)=1\\ \frac{2tg(\frac{x}{2} )}{1+tg^2(\frac{x}{2} )} +\frac{1-tg^2(\frac{x}{2} )}{1+tg^2(\frac{x}{2} )} =1\\tg(\frac{x}{2} )=t\\\frac{2t}{1+t^2} +\frac{1-t^2}{1+t^2} =1\\\frac{2t+1-t^2-1-t^2}{1+t^2} =0\\2t-2t^2=0\\2t(1-t)=0\\t=0\\t=1\\tg(\frac{x}{2} )=0\\\frac{x}{2} =\pi k\\x=2\pi k\\tg(\frac{x}{2} )=1\\\frac{x}{2} =\frac{\pi}{4} \\x=\frac{\pi}{2} +2\pi k\\x \neq \pi+2\pi k\\x=2\pi k\\x=\frac{\pi}{2} +2\pi k$
k∈Z