Question

If $x\ne \frac{k\pi }{2},k\in I$ and ${\left(\cos x\right)}^{{\sin }^{2}x-4\sin x+3}=1$, then all solutions of $x$ are given by

• A. $n\pi +{(-1)}^n\frac{\pi }{2};n\in I$
• B. $2n\pi \pm \frac{\pi }{2};n\in I$
• C. $(2n+1)\pi -\frac{\pi }{2};n\in I$
• D. None of these

Answer 1

The correct option is D None of these

${\left(\cos x\right)}^{{\sin }^{2}x-4\sin x+3}=1$
Case 1
$\cos x=1\Rightarrow x=2n\pi$
Case 2
${\sin }^{2}x-4\sin x+3=0\Rightarrow (\sin x-3)(\sin x-1)=0$
$\Rightarrow \sin x=3$ or $\sin x=1$
$\Rightarrow x=\frac{(4n+1)\pi }{2}$ as $\sin x\ne 3$