sangers1959

Verified answer

Sin⁴x+cos⁴x=5/8
sin⁴x+2sin²x*cos²x+cos⁴x-2sin²x*cos²x=5/8
(sin²x+cos²x)²-2sin²x*cos²x=5/8
1²-2sin²x*cos²x=5/8
2sin²x*cos²x=3/8
4sin²x*cos²x=3/4
(2sinx*cosx)²=3/4
sin²2x=3/4
sin2x=+/-√(3/4)=+/-√3/2
2x=+/-π/3+2πn
x=+/-π/6+πn
Ответ: x₁=π/6+πn        x₂=-π/6+πn.

oganesbagoyan

Verified answer

task/25172737 
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Решите уравнение sin⁴x +cos⁴x =5/8
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Решение
sin⁴x +cos⁴x =5/8 ⇔(sin²x +cos²x)² - 2*(sinxcosx)² =5/8 ⇔
1 -2*( (sin2x)/2 )²= 5/8⇔2*(sin²2x)/4 =1 - 5/8⇔(1/2)*sin²2x= 3/8⇔sin²2x= 3/4 ⇔(1-cos4x) /2 = 3/4 ⇔1-cos4x= 3/2 ⇔ cos4x = -1/2 .
4x =± (π -π/3) +2π*n , n ∈Z  ;
x =± π/6 +(π/2)*n , n ∈Z.

ответ : ± π/6 +(π/2)*n , n ∈Z.