Question

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

Answer 1

$\begin{array}{c}\sqrt{16{\cos }^{4}x-8{\cos }^{2}x+1}+\sqrt{16{\cos }^{4}x-24{\cos }^{2}x+9}=2\\ \sqrt{{\left(4{\cos }^{2}x-1\right)}^2}+\sqrt{{\left(4{\cos }^{2}x-3\right)}^2}=2\\ 4{\cos }^{2}x-1+4{\cos }^{2}x-3=2\\ 8{\cos }^{2}x-4=2\\ 8{\cos }^{2}x=6\\ {\cos }^{2}x=\frac{6}{8}\\ \cos x=\frac{\sqrt{3}}{2}\\ x={30}^0\\ x=\frac{\pi }{6}\end{array}$

Hence, this is the answer.