Question

How to solve sin3x+cos2x+2=0?

Answer 1

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

Answer 2

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.