Q65 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics
The solution set of the equation $\cos ^2 2 x+\sin ^2 3 x=1$ is
- A$\left\{x / x=n \pi+\frac{\pi}{2}, n \in z\right\}$
- B$\left\{x / x=2 n \pi \pm \frac{\pi}{4}, n \in z\right\}$
- Correct answerC$\left\{x / x=\frac{n \pi}{5}, n \in z\right\}$
- D$\left\{x / x=n \pi+(-1)^n \frac{\pi}{6}, n \in z\right\}$
Solution
$\cos ^2 2 x+\sin ^2 3 x=1$
$\begin{aligned}
& \Rightarrow \sin ^2 3 x=1-\cos ^2 2 x=\sin ^2 2 x \\
& \Rightarrow \sin ^2 3 x-\sin ^2 2 x=0 \\
& \Rightarrow \sin (3 x+2 x) \cdot \sin (3 x-2 x)=0 \\
& \Rightarrow \sin 5 x \cdot \sin x=0 \\
& \Rightarrow \sin 5 x=0 \Rightarrow 5 x=n \pi \Rightarrow x=\frac{n \pi}{5} \quad ...\mathrm{(i)} \\
& \Rightarrow \sin x=0 \Rightarrow x=n \pi \quad ...\mathrm{(ii)}
\end{aligned}$
From (i) and (ii) $x=\frac{n \pi}{5}, n \in Z$.
From (i) and (ii) $x=\frac{n \pi}{5}, n \in Z$.
Hence, the correct answer is option: 'C'.