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 \(\displaystyle\theta_{{{5}}}={3.393}+{2}{k}\pi\).
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 \(\displaystyle\theta_{{{5}}}={3.393}+{2}{k}\pi\).
