**Do you know how to solve the below question?**

\text { Solve } 1-\cos \left(\frac{x}{2}\right)=\cos \left(\frac{x}{2}\right) \text { for } x \in[-2 \pi, \pi]

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Still hesitated? No worries. Let’s take a look at the solution for the above trigonometric question first

Step 1: Conduct the algebraic manipulation to get .

Gives \cos \left(\frac{x}{2}\right)=\frac{1}{2}.

Step 2: Find the answer by using special values table, or by using calculator and inverse trigonometric.

From special values gives \frac{x}{2}=\frac{\pi}{3}.

Step 3: Apply symmetric properties.

\cos \left(-\frac{x}{2}\right)=\frac{1}{2}, so -\frac{x}{2}=\frac{\pi}{3}.

Step 4: Apply periodic properties.

Gives \frac{x}{2}=\frac{\pi}{3}+2 n \pi, x=\frac{2 \pi}{3}+4 n \pi. The other one gives x=-\frac{2 \pi}{3}+4 n \pi

Step 5: Conduct the necessary simplifications.

We have: x=\pm \frac{2 \pi}{3}+4 n \pi, n \in Z.

Step 6: Fit into the range of the questions.

Which is x=-\frac{2 \pi}{3}, \frac{2 \pi}{3}.

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