Question

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

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

Answer 1

Answer: (C)

$n\pi +\frac{\pi }{2}$

Given

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

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

The range of ${\sin }^{100}x$ is $[0,1]$ and $1+{\cos }^{100}x$ is $[1,2]$

The only possibility is ${\sin }^{100}x=1,{\cos }^{100}x=0$

$⟹x=n\pi +\frac{\pi }{2}$