Find the value of x∈[0,π]x \in [0, \pi]x∈[0,π] satisfying 2cos2(x)−sin(x)−1=02\cos^2(x) - \sin(x) - 1 = 02cos2(x)−sin(x)−1=0.
π/6,5π/6,π/2\pi/6, 5\pi/6, \pi/2π/6,5π/6,π/2
π/3,2π/3\pi/3, 2\pi/3π/3,2π/3
π/6,5π/6\pi/6, 5\pi/6π/6,5π/6
π/2\pi/2π/2