Question

If $\mid \cos x{\mid }^{{\sin }^{2}x-\frac{3}{2}\sin x+\frac{1}{2}}=1$, the possible values of x-

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

Answer 1

Answer: (A), (B), (D)

The correct options are
A $n\pi$ or $n\pi +(-1{)}^n\frac{\pi }{6},nϵI$
B $n\pi$ or $2n\pi +\frac{\pi }{2}$ or $n\pi +(-1{)}^n\frac{\pi }{6},nϵI$
D $n\pi ,nϵI$

$\mid \cos x\mid =1$ or ${\sin }^{2}x-\frac{3}{2}\sin x+\frac{1}{2}=0$
Either ${\sin }^{2}x-\frac{3}{2}\sin x+\frac{1}{2}=0$
or $|\cos x|=1$
${\sin }^{2}x-\frac{3}{2}\sin x+\frac{1}{2}=0$
$\Rightarrow \sin x=1$ or $\sin x=\frac{1}{2}$
$|\cos x|=1$
$\Rightarrow \cos x=-1$ or $\Rightarrow \cos x=1$
Hence, option A,B,D are the correct options