\section*{Exercise 2 - 031122}

Determine the set of all real numbers \(x\) that satisfy the following equation:

\[
1-\sin 5 x=\left(\cos \frac{3}{2} x-\sin \frac{3}{2} x\right)^{2}
\]
}

Apply the addition theorem

\[
\sin \alpha-\sin \beta=2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}
\]

to \(\alpha=\frac{\pi}{2}+\frac{3}{2} x\) and \(\beta=\frac{3}{2} x\) and obtain:

\[
\begin{aligned}
\cos \frac{3}{2} x-\sin \frac{3}{2} x & =\sin \left(\frac{\pi}{2}+\frac{3}{2} x\right)-\sin \left(\frac{3}{2} x\right) \\
& =2 \cos \frac{\pi+6 x}{4} \sin \frac{\pi}{4}=\sqrt{2} \cos \frac{\pi+6 x}{4} \\
\left(\cos \frac{3}{2} x-\sin \frac{3}{2} x\right)^{2} & =2 \cos ^{2} \frac{\pi+6 x}{4}=1+\cos \frac{\pi+6 x}{2} \\
& =1-\sin 3 x
\end{aligned}
\]

Thus, the original equation becomes

\[
1-\sin 5 x=1-\sin 3 x \Rightarrow \sin 5 x=\sin 3 x
\]

The left side is zero exactly when \(x\) is a multiple of \(\frac{\pi}{5}\) and the right side is zero exactly when \(x\) is a multiple of \(\frac{\pi}{3}\). The least common multiple of \(\frac{\pi}{5}\) and \(\frac{\pi}{3}\) is \(\pi\), so the equation is satisfied exactly when \(x\) is a multiple of \(\pi\).