Solve for x∈[0,π]x \in [0, \pi]x∈[0,π]: 4cos3x−3cosx=04\cos^3 x - 3\cos x = 04cos3x−3cosx=0.
π6,π2,5π6\frac{\pi}{6}, \frac{\pi}{2}, \frac{5\pi}{6}6π,2π,65π
π3,2π3\frac{\pi}{3}, \frac{2\pi}{3}3π,32π
π4,3π4\frac{\pi}{4}, \frac{3\pi}{4}4π,43π
0,π2,π0, \frac{\pi}{2}, \pi0,2π,π