Find the real zeros of the trigonometric function on the interval $0 \leq \theta< 2 \pi$.
\[
f(x)=-\sin (2 x)+\sin x
\]
Answer
\(\boxed{\text{Final Answer: The real zeros of the function } -\sin(2x) + \sin(x) \text{ on the interval } 0 \leq x < 2\pi \text{ are } 0, 1.04719755119660, 3.14159265358979, \text{ and } 5.23598775598299}\)
Step 1 ：We are given the function \(f(x) = -\sin(2x) + \sin(x)\) and we need to find the real zeros of this function in the interval \(0 \leq x < 2\pi\).
Step 2 ：The real zeros of a function are the x-values for which the function equals zero. So, we need to solve the equation \(-\sin(2x) + \sin(x) = 0\) for \(x\) in the interval \(0 \leq x < 2\pi\).
Step 3 ：Solving the equation gives us the values \(x = 0\), \(x = 1.04719755119660\), \(x = 3.14159265358979\), and \(x = 5.23598775598299\).
Step 4 ：\(\boxed{\text{Final Answer: The real zeros of the function } -\sin(2x) + \sin(x) \text{ on the interval } 0 \leq x < 2\pi \text{ are } 0, 1.04719755119660, 3.14159265358979, \text{ and } 5.23598775598299}\)
