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опять? task/29545472

362. 2)  √2(cos(π/4 -x) =(sinx+cosx)²⇔√2(cos(x -π/4) = [√2(cos(x -π/4) ] ²

2cos(x -π/4)* (cos(x - π/4 ) - 1/√2 ) =0 ⇔[ cos(x -π/4) =0; cos(x - π/4 ) =1/√2 .⇔

[ x -π/4  =π/2 +πn ;  x -π/4  =±π/4 +2πn ,n ∈ℤ .⇔

[ x=3π/4 +πn ; x =2πn ; x=π/2+  2πn, n ∈ℤ .

ответ : π/4 +πn ;  2πn;  π/2+  2πn, n ∈ℤ .

363. 4)  sin³xcosx +cos³xsinx = 1/4 ⇔ sinx*cosx(sin²x +cos²x) = 1/4 ⇔

sinx*cosx = 1/4 ⇔(1/2)sin2x = 1/4 ⇔sin2x = 1/2 ⇒ 2x =(-1)ⁿπ/6+πn , n ∈ℤ  .

ответ : x = (-1)ⁿ*π/12 +(π/2)n , n ∈ℤ  .