√ 7-8sinx = - 2cosx при x∈ (-3π/2 ;3π/2 ).
ОДЗ : {sinx≤ 7/8 ;cosx≤ 0.
7-8sinx = 4cos²x ⇔4sin²x -8sinx+3 =0 ⇒ sinx₁ =3/2 (не имеет решения) и
sinx =1/2 ; [ 7-8sinx=7-8*1/2 =3 ≥0 ] .
cosx = - √3/2 ;
x₁= 5π/6 +2π*k ; -3π/2 <5π/6 +2π*k<3π/2⇒ -7/6 <k <1/3. k =-1;0.<br>x₁= -7π/6 , x₁= 5π/6
x₂ = - 5π/6+2π*n ; -3π/2 <-5π/6 +2π*k<3π/2 ⇒ -1/3 <n <7/6 .n=0 ;1.<br>x₂ = 7π/6. x₂ = -5π/6.