Question

If ${\sin }^{3}x+{\cos }^{3}x+\sin x\cos x=1$, then $x$ is equal to

• A. $2n\pi ,nϵZ$
• B. $(2n+1)\pi ,nϵZ$
• C. $2n\pi +\frac{\pi }{2},nϵZ$
• D. $2n\pi -\frac{\pi }{2},nϵZ$

Answer 1

Answer: (A), (C)

The correct options are
A $2n\pi ,nϵZ$
C $2n\pi +\frac{\pi }{2},nϵZ$

Given equation can be written as $(\sin x+\cos x)({\sin }^{2}x-\sin x\cos x+{\cos }^{2}x)-(1-\sin x\cos x)=0$

$\Rightarrow (\sin x+\cos x)(1-\sin x\cos x)-(1-\sin xcosx)=0$

$\Rightarrow (\sin x+\cos x-1)(1-\sin x\cos x)=0$

$\Rightarrow \sin x+\cos x=1$

or

$1-\sin x\cos x=0$

This gives

$\cos (x-\frac{\pi }{4})=\frac{1}{\sqrt{2}}$

or

$\sin 2x=2$

But $\sin 2x$ can't be equal to 2.

Therefore,

$\cos (x-\frac{\pi }{4})=\frac{1}{\sqrt{2}}$

$\Rightarrow x-\frac{\pi }{4}=2n\pi \pm \frac{\pi }{4}$

$\Rightarrow x=2n\pi +\frac{\pi }{2}$

or

$x=2n\pi ,nϵZ$