ответ : x = ±π/2 +(3π/2)*n , n∈ z . -------.--- P.S.-------.--- sin2α= 2sinα*cosα ⇒ sinα*cosα =(1/2)sin2α ⇔sin²α*cos²α =(1/4)sin²2α
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oganesbagoyan
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)
Answers & Comments
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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
Verified answer
(1-cos2x/3)²/4+(1+cos2x/3)²/4=5/81-2cos2x/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