The given equation (\tan \theta - 2) (16 \sin^{2} \theta - 1) = 0

Approach:
The domain of the trigonometry function of ${\displaystyle \tan \theta }$ lies between ${\displaystyle [-\mathrm{\infty },\mathrm{\infty }]}$. Tangent has period ${\displaystyle \pi }$, we findd solution in any interval of length ${\displaystyle \pi }$.
The domain of the trigonometric function ${\displaystyle \sin \theta }$ lies between ${\displaystyle [-1,1]}$. No solution exists beyond this domain. Sine has period ${\displaystyle 2\pi }$, we find solution in any interval of length ${\displaystyle 2\pi }$. Sine function is positive in first and second quadrant.
Calculation:
The trigonometric equation is given by,
${\displaystyle (\tan \theta -2)(16{\sin }^2\theta -1)=0}$
The factors of above equation are,
${\displaystyle (\tan \theta -2)=0\dots \left(1\right)}$
${\displaystyle 16{\sin }^2\theta -1=0\dots \left(2\right)}$
Consider the factors.
Add 2 both sides in equation (1).
${\displaystyle \tan \theta =2}$
${\displaystyle \tan \theta ={\tan }^{-1}\left(2\right)}$
${\displaystyle \theta =1.10714}$
Here the angles are in radian.
The tangent has period, ${\displaystyle \pi }$, so we get all solutions of the equation by adding integer multiples of ${\displaystyle \pi }$ to these solutions:
${\displaystyle {\theta }_1=1.10714+k\pi }$
The solution obtained for the factor in which sine function involved so we will get the solution in the interval of ${\displaystyle [0,2\pi ]}$.
Add 1 to both sides in equation (2).
${\displaystyle 16{\sin }^2\theta =1}$
${\displaystyle \sin \theta =\pm \frac{1}{4}}$
${\displaystyle \theta ={\sin }^{-1}(\pm \frac{1}{4})}$
Taking positive sign,
${\displaystyle \theta ={\sin }^{-1}\left(\frac{1}{4}\right)}$
${\displaystyle =0.252,2.889}$
Taking negative sign,
${\displaystyle \theta ={\sin }^{-1}\left(\frac{1}{4}\right)}$
${\displaystyle =-0.252,3.393}$
The sine has period, ${\displaystyle 2\pi }$, but it is a square function so repeating is done every length of ${\displaystyle \pi }$. So we get all solutions of the equation by adding integer multiples of ${\displaystyle \pi }$ to these solutions:
${\displaystyle {\theta }_2=0.252+2k\pi }$
${\displaystyle {\theta }_3=2.889+2k\pi }$
${\displaystyle {\theta }_4=-0.252+2k\pi }$
${\displaystyle {\theta }_5=3.393+2k\pi }$
Therefore, the solutions of the trigonometric equation ${\displaystyle (\tan \theta -2)(16{\sin }^2\theta -1)=0}$ are ${\displaystyle {\theta }_1=1.10714+k\pi ,{\theta }_2=0.252+2k\pi ,{\theta }_3=2.889+2k\pi ,{\theta }_4=-0.252+2k\pi }$ and ${\displaystyle {\theta }_5=3.393+2k\pi }$.