Sin в 4 степениx/3 плюс cos в 4 степениx/3=5/8
Task/25178816 * * * t =x/3 , sin⁴t +cos⁴t=5/8 → task/25172737 * * * ---------------------- sin⁴(x/3) +cos⁴(x/3)=5/8 ; ( sin²(x/3) +cos²(x/3) )² -2sin²(x/3) *cos²(x/3) = 5/8 ; 1 - (1/2)sin²(2x/3) =5/8 ; sin²(2x/3) =3/4 ; * * * cos²(2x/3) =1-3/4 =1/4 * * * ( 1 -cos(4x/3) ) /2 = 3/4 ; cos(4x/3) = -1/2 ; * * * cos4t = -1/2 * * * 4x/3 = ±(π- π/3) +2πn , n∈ z ; x = ±π/2 +(3π/2)*n , n∈ z . ответ : x = ±π/2 +(3π/2)*n , n∈ z . -------.--- P.S.-------.--- sin2α= 2sinα*cosα ⇒ sinα*cosα =(1/2)sin2α ⇔sin²α*cos²α =(1/4)sin²2α
(1-cos2x/3)²/4+(1+cos2x/3)²/4=5/8 1-2cos2x/3+cos²2x/3+1+2cos2x/3+cos²2x/3=5/2 2cos²2x/3+2=2,5 2cos²2x/4=0,5 cos²2x/3=0,25 cos2x/3=-0,5 U cos2x/3=0,5 2x/3=+-5π/6+2πk U x=+-π/6+2πk,k∈z
последняя строка 2x/3=+-5π/6+2πk U x=+-π/6+2πk,k∈z неверно , заменить на 2x/3=±2π/3+2πk U 2x/3=±π/3+2πk,k∈z , а затем добавить x=±π+3πk U x=±π/2+3πk,k∈z (ответ)