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task/25178816 * * * t =x/3 , sin⁴t +cos⁴t=5/8 → task/25172737 * * *
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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
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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
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