Question

Find the general solution of the trigonometric equation :
$\sqrt{16{\cos }^{4}x-8{\cos }^{2}x+1}+\sqrt{16{\cos }^{4}x-24{\cos }^{2}x+9}=2$.

• A. $x\in [n\pi +\frac{\pi }{6},n\pi +\frac{\pi }{3}]\cup [n\pi +\frac{2\pi }{3},n\pi +\frac{5\pi }{6}],n\in I$
• B. $x\in [n\pi +\frac{\pi }{3},n\pi +\frac{\pi }{3}]\cup [n\pi +\frac{2\pi }{3},n\pi +\frac{5\pi }{6}],n\in I$
• C. $x\in [n\pi +\frac{\pi }{4},n\pi +\frac{\pi }{3}]\cup [n\pi +\frac{2\pi }{3},n\pi +\frac{5\pi }{6}],n\in I$
• D. $x\in [n\pi +\frac{\pi }{5},n\pi +\frac{\pi }{3}]\cup [n\pi +\frac{2\pi }{3},n\pi +\frac{5\pi }{6}],n\in I$

Answer 1

The correct option is A $x\in [n\pi +\frac{\pi }{6},n\pi +\frac{\pi }{3}]\cup [n\pi +\frac{2\pi }{3},n\pi +\frac{5\pi }{6}],n\in I$

$\sqrt{16{\cos }^{4}x-8{\cos }^{2}x+1}+\sqrt{16{\cos }^{4}x-24{\cos }^{2}x+9}=2\sqrt{(4{\cos }^{2}x-1{)}^2}+\sqrt{(4{\cos }^{2}x-3{)}^2}=2|4{\cos }^{2}x-1|+|4{\cos }^{2}x-3|=2Case-I:If{\cos }^{2}x<\frac{1}{4},then,1-4{\cos }^{2}x+3-4{\cos }^{2}x=2\Rightarrow 8{\cos }^{2}x=2{\cos }^{2}x=\frac{1}{4}\left(rejected\right)Case-II:If,\frac{1}{4}\le {\cos }^{2}x\le \frac{3}{4}then,4{\cos }^{2}x-1+3-4{\cos }^{2}x=22=2,\cos xϵ[\frac{-\sqrt{3}}{2},-\frac{1}{2}]U[\frac{1}{2},\frac{\sqrt{3}}{2}]\Rightarrow xϵ[n\pi +\frac{\pi }{6},n\pi +\frac{\pi }{3}]U[n\pi +\frac{2\pi }{3},n\pi +\frac{5\pi }{6}]nϵICase-III:If,{\cos }^{2}x>\frac{3}{4},then,4{\cos }^{2}x-1+4{\cos }^{2}x-3=2{\cos }^{2}x=\frac{3}{4}\left(rejected\right)$