Question

Решить уравнение:4sin^2 х+cosx-3=0

Answer 1

1-4cos^2(x)-cos(x)=0                                                                                   
cos(x)=y                                                                                                      
y^2+(2/8)x=1/4  y^2+(2/8)x+1/64=17/64                                                                          
y1=-1/8(1-sqrt(17))                                                                                                                       y2=-1/8(1+sqrt(17))                                                                                                                         
х1=arccos(-1/8(1-sqrt(17)))+2pi*k                                                               
x2=arccos(1/8(1-sqrt(17)))+2pi*k                                                     
х1=arccos(-1/8(1+sqrt(17)))+2pi*k                                                            
х1=arccos(1/8(1+sqrt(17)))+2pi*k                                              
k-любое целое