How to solve sin3x+cos2x+2=0?
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.
Hello,sin (3x)=3sin(x)-4sin^3(x)cos(2x)=1-2sin²(x)Thus -4sin ^3(x)-2sin²(x)+3sin(x)+3=0Lest's assume y=sin (x)4y^3+2y²-3y-3=0==>(y-1)(4y²+6y+3)=0Only one real solution: sin(x)=1==>x=π/2+2kπ
