Sqrt 2 sin^2(pi/2+x)=-cosx
√2 sin²(π/2+x)=-cosx, √2 sin²(π/2+x)=√2сos²x=-cosx √2cosx=-cosx -√2 cosx =-cosx (√2+1) cosx =0 (1-√2) cosx =0 х=π/2+2πn, n∈Z