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 .