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

Factors and multiples
ANSWERED The given equation $$\displaystyle{\left({\tan{\theta}}-{2}\right)}{\left({16}{{\sin}^{{{2}}}\theta}-{1}\right)}={0}$$ 2021-08-12
Approach:
The domain of the trigonometry function of $$\displaystyle{\tan{\theta}}$$ lies between $$\displaystyle{\left[-\infty,\infty\right]}$$. 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 $$\displaystyle{\left[-{1},{1}\right]}$$. 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,
$$\displaystyle{\left({\tan{\theta}}-{2}\right)}{\left({16}{{\sin}^{{{2}}}\theta}-{1}\right)}={0}$$
The factors of above equation are,
$$\displaystyle{\left({\tan{\theta}}-{2}\right)}={0}\ldots{\left({1}\right)}$$
$$\displaystyle{16}{{\sin}^{{{2}}}\theta}-{1}={0}\ldots{\left({2}\right)}$$
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 $$\displaystyle{\left[{0},{2}\pi\right]}$$.
Add 1 to both sides in equation (2).
$$\displaystyle{16}{{\sin}^{{{2}}}\theta}={1}$$
$$\displaystyle{\sin{\theta}}=\pm{\frac{{{1}}}{{{4}}}}$$
$$\displaystyle\theta={{\sin}^{{-{1}}}{\left(\pm{\frac{{{1}}}{{{4}}}}\right)}}$$
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:
$$\displaystyle\theta_{{{2}}}={0.252}+{2}{k}\pi$$
$$\displaystyle\theta_{{{3}}}={2.889}+{2}{k}\pi$$
$$\displaystyle\theta_{{{4}}}=-{0.252}+{2}{k}\pi$$
$$\displaystyle\theta_{{{5}}}={3.393}+{2}{k}\pi$$
Therefore, the solutions of the trigonometric equation $$\displaystyle{\left({\tan{\theta}}-{2}\right)}{\left({16}{{\sin}^{{{2}}}\theta}-{1}\right)}={0}$$ are $$\displaystyle\theta_{{{1}}}={1.10714}+{k}\pi,\theta_{{{2}}}={0.252}+{2}{k}\pi,\theta_{{{3}}}={2.889}+{2}{k}\pi,\theta_{{{4}}}=-{0.252}+{2}{k}\pi$$ and $$\displaystyle\theta_{{{5}}}={3.393}+{2}{k}\pi$$.