Question

The general solution of the equation ${\sin }^{100}x-{\cos }^{100}x=1$

• A. $2n\pi +\frac{\pi }{3},n\in I$
• B. $n\pi +\frac{\pi }{2},n\in I$
• C. $n\pi +\frac{\pi }{4},n\in I$
• D. $2n\pi -\frac{\pi }{3},n\in I$

Answer 1

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

${\sin }^{100}x-{\cos }^{100}x=1$

$\Rightarrow {\sin }^{100}x=1+{\cos }^{100}x$

Now, maximum value of ${\sin }^{100}x$ can be $1$

So,above equation will satisfy only when ${\cos }^{100}x=0$

$\therefore {\sin }^{100}x=1$ and ${\cos }^{100}x=0$ is the only solution for this equation.

Now ${\sin }^{100}x=1$

$\Rightarrow {\sin }^{2}x=1$

$\Rightarrow \sin x=\pm 1$

$\therefore x=n\pi +\frac{\pi }{2}$ is the general solution.