Question

Решите уравнение а) √2sinx=2-√2cosx б) 2(tgx/2-1)=cosx

Answer 1

$a) \sqrt{2}sinx=2- \sqrt{2}cosx; \\ \sqrt{2}sinx + \sqrt{2}cosx =2; ~|:2 \\ \frac{ \sqrt{2}}{2} sinx+ \frac{ \sqrt{2}}{2} cosx=1; \\ cos \frac{ \pi }{4} sinx+sin \frac{ \pi }{4} cosx =1; \\ sin( \frac{ \pi }{4} +x)=1; \\ \frac{ \pi }{4} +x= \frac{ \pi }{2}+2 \pi n,~n\in Z \\ x= \frac{ \pi }{4}+2 \pi n,~n\in Z \\ b) Oshibka~v~uslovii?$