Find the real zeros of the trigonometric function on the interval \( 0 \leq \theta<2 \pi \). \( f f(x)=6 \boldsymbol{\operatorname { s i n }}(2 x)-6 \boldsymbol{\operatorname { s i n }} x \) The real zeros are (Use a comma to separate answers as needed. Type exact answers, using \( \pi \) as needed. Use integers or fractions for any numbers in the expressions.)

To find the real zeros of the function \( f(x) = 6 \sin(2x) - 6 \sin(x) \), we can start by setting this equal to zero: \[ 6 \sin(2x) - 6 \sin(x) = 0 \] Dividing by 6 gives us: \[ \sin(2x) - \sin(x) = 0 \] Using the double angle identity, \( \sin(2x) = 2 \sin(x) \cos(x) \), we can rewrite the equation: \[ 2 \sin(x) \cos(x) - \sin(x) = 0 \] Factoring out \( \sin(x) \): \[ \sin(x)(2 \cos(x) - 1) = 0 \] Setting each factor to zero gives us two cases: 1. \( \sin(x) = 0 \) 2. \( 2 \cos(x) - 1 = 0 \) For \( \sin(x) = 0 \), we find: \[ x = 0, \pi, 2\pi \] However, since we are looking in the interval \( 0 \leq x < 2\pi \), the relevant solutions are: \[ x = 0, \pi \] For \( 2 \cos(x) - 1 = 0 \), we solve for \( \cos(x) = \frac{1}{2} \): \[ x = \frac{\pi}{3}, \frac{5\pi}{3} \] Combining the results, the real zeros of the function \( f(x) \) in the interval \( 0 \leq x < 2\pi \) are: \[ 0, \pi, \frac{\pi}{3}, \frac{5\pi}{3} \]