Question

2sin^2x+14cosx-3cos^2x=0

Answer 1

2sin²x +14cosx -3cos²x =0 ;
2(1-cos²x) +14cosx -3cos²x =0 ;
5cos²x -14cosx -2 =0 ; 
* * * замена :  t =cosx ; | t | ≤1 * * *
5t² -14t -2 =0 ;
t₁ =(7 +√59)/5 >1 не решение 
t₂ =(7 -√59)/5.
cosx = (7 -√59)/5 ;

x = ±arccos (7 -√59)/5 +2π*n , n∈Z.