Q65 / 155 · WBJEE 2026 · Mathematics
The general solution of the equation $\sin^{100} x - \cos^{100} x = 1$ is
- A$\left\{2n\pi + \dfrac{\pi}{3} : n \in I\right\}$
- B$\left\{n\pi + \dfrac{\pi}{4} : n \in I\right\}$
- Correct answerC$\left\{n\pi \pm \dfrac{\pi}{2} : n \in I\right\}$
- D$\left\{2n\pi - \dfrac{\pi}{3} : n \in I\right\}$
Solution
The given equation is $\sin^{100} x - \cos^{100} x = 1$.
We know that for any real $x$, $\sin^{100} x \le 1$ and $\cos^{100} x \ge 0$.
Therefore, $\sin^{100} x - \cos^{100} x \le 1$.
The equality holds if and only if $\sin^{100} x = 1$ and $\cos^{100} x = 0$.
$\sin^{100} x = 1 \Rightarrow \sin^2 x = 1 \Rightarrow \cos^2 x = 0$.
This gives $\cos x = 0$.
The general solution for $\cos x = 0$ is $x = n\pi \pm \dfrac{\pi}{2}$, where $n \in I$.
Thus, the solution set is $\left\{n\pi \pm \dfrac{\pi}{2} : n \in I\right\}$.
Answer: $\left\{n\pi \pm \dfrac{\pi}{2} : n \in I\right\}$
We know that for any real $x$, $\sin^{100} x \le 1$ and $\cos^{100} x \ge 0$.
Therefore, $\sin^{100} x - \cos^{100} x \le 1$.
The equality holds if and only if $\sin^{100} x = 1$ and $\cos^{100} x = 0$.
$\sin^{100} x = 1 \Rightarrow \sin^2 x = 1 \Rightarrow \cos^2 x = 0$.
This gives $\cos x = 0$.
The general solution for $\cos x = 0$ is $x = n\pi \pm \dfrac{\pi}{2}$, where $n \in I$.
Thus, the solution set is $\left\{n\pi \pm \dfrac{\pi}{2} : n \in I\right\}$.
Answer: $\left\{n\pi \pm \dfrac{\pi}{2} : n \in I\right\}$
Hence, the correct answer is option: 'C'.