Task/26481635
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2sin²x +1/ cos²x =3 ; * * * ОДЗ : cosx ≠ 0 * * *
2(1 -cos²x)*cos²x +1 = 3cos²x ;
2cos²x -2cos⁴x +1 - 3cos²x =0 ;
2cos⁴x +cos²x - 1 = 0 ; замена : t =cos²x ≥ 0
2t² +t -1 = 0 ; * * * t = -1 * * *
t₁ =( -1 -3)/4 = -1 < 0 → посторонний корень
t₂ = (-1+3)/4 = 1/2 .⇒ cos²x = 1/2 ⇔ (1+cos2x) / 2 =1/2 ⇔ 1+cos2x =1 ⇔
cos2x =0 ;
2x =π/2 +πn , n ∈ Z .
x = π/4 +(π/2)*n , n ∈ Z .
ответ : π/4 +(π/2)*n , n ∈ Z .