Question

8sin^2x+6sin(pi/2+x)=9

Answer 1

надеюсь все видно....

Answer 2

sin(π/2 + x) = cosx
8 (1-cos²x) + 6cosx = 9
8- 8cos²x + 6cosx -9=0
-8cos²x + 6cosx - 1=0
8cos²x - 6cosx + 1=0
cosx = t;  ItI≤1
8t² - 6t + 1=0
D₁ = 9-8=1
t₁ = (3+1)/8 = 1/2
t₂ = (3-1)/8 = 1/4
cosx = 1/2            cosx = 1/4
x = π/3 + 2πn          x = arccos1/4  + 2πn