4) cos3x +sinxsin2x =2cos³x +tqx ;
cos3x +sinxsin2x =2cos²x*cosx +tqx ;
cos3x +sinxsin2x =(1+cos2x)*cosx +tqx ;
cos3x - (cosxcos2x -sinxsin2x) =cosx +tqx ;
cos3x -cos3x =cosx +sinx/cosx ;
(cos²x+sinx)/cosx = 0;
{ cos²x +sinx =0 ; cosx ≠0;
1 - sin²x +sinx =0 ;
sin²x -sinx -1 =0 ;
a) sinx = (1+√5)/2 >1_ нет решения ;
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b) sinx = (1 -√5)/2 ;
x =(-1)^(n+1) arcsin(√5 -1)/2 +π*n , n∈Z.
5) 2cos²x + 3sinx = 0 ;
2(1-sin²x) +3sinx = 0;
2sin²x -3sinx -2 =0;
[sinx =2 ;sinx= - 1/2.
sinx =2>1_ нет решения ;
sinx= - 1/2.
x =(1)^(n+1)π/6 +π*n , n ∈Z .
6) 2sin²x + 1/cos²x = 3 ;
замена переменного : t =sin²x , 0≤ t ≤1 .
2t +1/(1-t) =3;
2t² - 5t +2 =0;
t₁ =2_ нет решения ;
t₂=1/2;
sin²x =1/2;
(1-cos2x)/2 =1/2;
cos2x =0;
2x =2π*n ;n∈Z.
ответ :x = π*n , n∈Z.
7) sin3x -sinx +cos2x =1 ;
2sin(3x-x)/2*cos(3x+x)/2 -(1 -cos2x) = 0;
2sinxcos2x - 2sin²x =0;
2sinx(cos2x -sinx) =0 ; *** [ sinx = 0 ; cos2x -sinx =0 ] ***
a) sinx =0 ⇒x =π*n , n∈Z.
b) cos2x -sinx) =0 ;
1 -2sin²x -sinx =0;
2sin²x +sinx -1 =0 ;
[sinx = - 1 ; sinx =1/2 . [x = -π/2+2π*n ; x = (-1)^(n)π/6 +π*n ,n ∈Z.
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ответ : π*n ; -π/2+2π*n ; (-1)^(n)π/6 +π*n ,n ∈Z.