Question

The general solution of the equation ${\sin }^{100}x-{\cos }^{100}x=1$ is
(where $n\in Z$ )

• A. $2n\pi +\frac{\pi }{2}$
• B. $n\pi +\frac{\pi }{2}$
• C. $2n\pi -\frac{\pi }{2}$
• D. $n\pi$

Answer 1

The correct option is B $n\pi +\frac{\pi }{2}$

${\sin }^{100}x-{\cos }^{100}x=1$
$\Rightarrow {\sin }^{100}x=1+{\cos }^{100}x$
We know that,
${\sin }^{100}x\le 1\text{and}1+{\cos }^{100}x\ge 1$
The equation has solution only when
$\Rightarrow {\sin }^{100}x=1\text{and}1+{\cos }^{100}x=1\Rightarrow {\sin }^{2}x=1\text{and}{\cos }^{2}x=0$
$\Rightarrow x=n\pi \pm \frac{\pi }{2}\text{and}x=n\pi \pm \frac{\pi }{2}$
$\Rightarrow x=n\pi \pm \frac{\pi }{2},n\in Z\Rightarrow x=\frac{(2n\pm 1)\pi }{2}$
As $(2n+1)$ and $(2n-1)$ both represent set of odd integers, so
$x=\frac{(2n+1)\pi }{2}=n\pi +\frac{\pi }{2},n\in Z$