Question

Solve the equation: $\sqrt{(\frac{1}{16}+co{s}^{4}x-\frac{1}{2}co{s}^{2}x)}+\sqrt{(\frac{9}{16}+co{s}^{4}x-\frac{3}{2}co{s}^{2}x)}=\frac{1}{2}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

The given equation
$\sqrt{(\frac{1}{16}+co{s}^{4}x-\frac{1}{2}co{s}^{2}x)}+\sqrt{(\frac{9}{16}+co{s}^{4}x-\frac{3}{2}co{s}^{2}x)}=\frac{1}{2}$
or, |cos^2x-{1\over 4}|+|cos^2x-{3\over 4}|={1\over 2}\)
Case I: If $co{s}^{2}x\ge \frac{3}{4}$
then $co{s}^{2}x-\frac{1}{4}+co{s}^{2}x-\frac{3}{4}=\frac{1}{2}$
or $2co{s}^{2}x-1=\frac{1}{2}$
or $cos2x=\frac{1}{2}=cos\frac{\pi }{3}\therefore 2x=2n\pi \pm \frac{\pi }{3}$
$\therefore x=n\pi \pm \frac{\pi }{6},n=0,\pm 1,\pm 2,........$ .....(1)
Case II: If $\frac{1}{4}\le co{s}^{2}x<\frac{3}{4}$
then $co{s}^{2}x-\frac{1}{4}+\frac{3}{4}-co{s}^{2}x=\frac{1}{2}\Rightarrow \frac{1}{2}=\frac{1}{2}$
$\therefore \frac{1}{4}\le co{s}^{2}x<\frac{3}{4}$
or, $\frac{1}{2}\le 2co{s}^{2}x<\frac{3}{2}or,-\frac{1}{2}\le cos2x<\frac{1}{2}$
or, $cos\frac{2\pi }{3}\le cos2x<cos\frac{\pi }{3}$
or, $2n\pi \pm \frac{2\pi }{3}\ge 2x>2n\pi \pm \frac{\pi }{3}$
$\therefore n\pi +\frac{\pi }{6}<x\le n\pi +\frac{\pi }{3}$ and
$n\pi -\frac{\pi }{3}\ge x>n\pi -\frac{\pi }{6}$ ....(2)
Hence solution from (1) and (2) is
$n\pi +\frac{\pi }{6}\le x\le n\pi +\frac{\pi }{3}$ and
$n\pi -\frac{\pi }{3}\le x\le n\pi -\frac{\pi }{6}$ where $n\in I$.