Solve 2cos x+2cos 2x=0 on the interval [0,2pi)
\(2cosx+2cos2x=0 \\ -2sin^2x+2cos^2x+2cosx=0 \\ -2(1-cos^2x)+2cos^2x+2cosx=0 \\ -2+2cos^2x+2cos^2x+2cosx=0 \\ 4cos^2x+2cosx-2=0 \\ t=cos x \\ 4t^2+2t-2=0 \\ \Delta=4+32=36 \\ t_1= \frac{-2-6}{8}=-1 \\ t_2= \frac{-2+6}{8}= \frac{1}{2} \\ cos x=-1 \lor cos x= \frac{1}{2} \\ x=-\pi \lor x= \frac{5\pi}{3}\)
Answer:
The answers are B and C, 3π/4 and 5π/4
Step-by-step explanation:
