Question

Solve 2cos x+2cos 2x=0 on the interval [0,2pi)

Answer 1

`2cosx+2cos2x=0 \\ -2sin^2x+2cos^2x+2cosx=0 \\ -2(1-cos^2x)+2cos^2x+2cosx=0 \\ -2+2cos^2x+2cos^2x+2cosx=0 \\ 4cos^2x+2cosx-2=0 \\ t=cos x \\ 4t^2+2t-2=0 \\ \Delta=4+32=36 \\ t_1= (-2-6)/(8)=-1 \\ t_2= (-2+6)/(8)= (1)/(2) \\ cos x=-1 \lor cos x= (1)/(2) \\ x=-\pi \lor x= (5\pi)/(3)`