Find the value of xxx in the interval [0,2π][0, 2\pi][0,2π] that satisfies the equation 2sin2(x)+3cos(x)−3=02\sin^2(x) + 3\cos(x) - 3 = 02sin2(x)+3cos(x)−3=0.
x=π3x = \frac{\pi}{3}x=3π
x=5π3x = \frac{5\pi}{3}x=35π
x=0x = 0x=0
x=πx = \pix=π