Question

how do i solve the equation cos(x)+2cos(x)sin(x)=0 on the interval [0,2π)?

Answer 1

Pull out a common factor:
cosx + 2 cosx sinx = 0
cosx (1 + 2 sinx) = 0

The zero product property gives you two equations:
cosx = 0
1 + 2 sinx = 0, or sinx = -1/2

Solve for x in the given interval.

Answer 2

also this is a solution
\[\cos(x)+2\cos(x)\sin(x)=0 \\cos(x)=-2\cos(x)\sin(x)\\cos(x)=-\sin(2x)\\cos(x)=\sin(-2x)\\cos(x)=\cos(\frac{ \pi }{ 2 }-(-2x))\\cos(x)=\cos(\frac{ \pi }{ 2 }+2x)\\x=\pm (\frac{ \pi }{ 2 }+2x)+ 2\pi k\]

Answer 3

thanks!!!