2sinxcosx+√2sinx=2cosx+√2
√2sinx(√2cosx+1)-√2(√2cosx+1)=0
(√2cosx+1)(√2sinx-√2)=0
cosx=-1/√2⇒x=+-3π/4+2πn,n∈Z
sinx=1⇒x=π/2+2πk,k∈Z
1)π≤-3π/4+2πn≤5π/2
4≤-3+8n≤10
7≤8n≤13
7/8≤n≤13/8
n=1⇒x=-3π/4+2π=5π/4
2)π≤3π/4+2πn≤5π/2
4≤3+8n≤10
1≤8n≤7
1/8≤n≤7/8
нет решения
3)π≤π/2+2πk≤5π/2
2≤1+4k≤5
1≤4k≤4
1/4≤k≤1
k=1⇒x=π/2+2π=5π/2