Question

The solution set of the equation \(\cos^2(2x) + \sin^2(3x) = 1\) is \;?

• A. \(\bigl\{\,x \mid x = n\pi + \tfrac{\pi}{2},\,n \in \mathbb{Z}\bigr\}\)
• B. \(\bigl\{\,x \mid x = 2n\pi \pm \tfrac{\pi}{4},\,n \in \mathbb{Z}\bigr\}\)
• C. \(\bigl\{\,x \mid x = \tfrac{n\pi}{5},\,n \in \mathbb{Z}\bigr\}\)
• D. \(\bigl\{\,x \mid x = n\pi + (-1)^n \tfrac{\pi}{6},\,n\in \mathbb{Z}\bigr\}\)

Hint:

When \(\cos^2A +\sin^2B=1\), see if it implies \(\cos^2A=\cos^2B\).
- Solve \(\cos p=\pm \cos q\) via standard angle-equality conditions.

Answer 1

The Correct Option is C

Step 1: Observe identity structure.
\[ \cos^2(2x) + \sin^2(3x) = 1. \] Recall \(\sin^2(\theta)=1-\cos^2(\theta)\). So the equation suggests \(\cos^2(2x) = \cos^2(3x)\). Hence \[ \cos^2(2x) - \cos^2(3x) = 0. \] Step 2: Factor the difference of squares.
\[ \cos^2(2x) - \cos^2(3x) = \bigl[\cos(2x)-\cos(3x)\bigr]\bigl[\cos(2x)+\cos(3x)\bigr] =0. \] So either \[ \cos(2x)=\cos(3x) \quad\text{or}\quad \cos(2x)=-\cos(3x). \] Step 3: Solve each case.
- \(\cos(2x)=\cos(3x)\) implies \(2x = 3x + 2n\pi\) or \(2x = -3x + 2m\pi\).
- \(\cos(2x)=-\cos(3x)\) implies \(2x = \pi \pm 3x + 2k\pi\), etc. A careful solution reduces to solutions of the form \(x=\frac{n\pi}{5}\), \(n\in\mathbb{Z}\).