Respuesta :

0100010
It depends on which field this equation is defined. Here you have some solution for real and the beginning of calculations defined in complex field.

Ver imagen 0100010
caylus
Hello,

sin (3x)=3sin(x)-4sin^3(x)
cos(2x)=1-2sin²(x)

Thus -4sin ^3(x)-2sin²(x)+3sin(x)+3=0
Lest's assume y=sin (x)

4y^3+2y²-3y-3=0
==>(y-1)(4y²+6y+3)=0
Only one real solution: sin(x)=1==>x=π/2+2kπ