Question

If ${\left|\cos x\right|}^{{\sin }^{2}x-\frac{3}{2}\sin x+\frac{1}{2}}=1$, then possible values of $x$ are:

• A. $n\pi$ or $n\pi +{(-1)}^n\frac{\pi }{6},n\in I$
• B. $n\pi$ or $2n\pi +\frac{\pi }{2}$ or $n\pi +{(-1)}^n\frac{\pi }{6},n\in I$
• C. $n\pi +{(-1)}^n\frac{\pi }{6},n\in I$
• D. $n\pi ,n\in I$

Answer 1

Answer: (C), (D)

The correct options are
C $n\pi +{(-1)}^n\frac{\pi }{6},n\in I$
D $n\pi ,n\in I$

The equation holds if $\left|\cos x\right|=1$

i.e. if $x=n\pi ,n\in I$

If $\left|\cos x\right|\ne 1$ then ${\sin }^{2}x-\frac{3}{2}\sin x+\frac{1}{2}=0$

$\Rightarrow \sin x=1$ or $\frac{1}{2}$

But $\sin x\ne 1$ as in that case $\cos x=0$

$\therefore \sin x=\frac{1}{2}\Rightarrow x=n\pi +{(-1)}^n\frac{\pi }{6}$