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

The given equation $\left(\mathrm{tan}\theta -2\right)\left(16{\mathrm{sin}}^{2}\theta -1\right)=0$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

l1koV
Approach:
The domain of the trigonometry function of $\mathrm{tan}\theta$ lies between $\left[-\mathrm{\infty },\mathrm{\infty }\right]$. Tangent has period $\pi$, we findd solution in any interval of length $\pi$.
The domain of the trigonometric function $\mathrm{sin}\theta$ lies between $\left[-1,1\right]$. No solution exists beyond this domain. Sine has period $2\pi$, we find solution in any interval of length $2\pi$. Sine function is positive in first and second quadrant.
Calculation:
The trigonometric equation is given by,
$\left(\mathrm{tan}\theta -2\right)\left(16{\mathrm{sin}}^{2}\theta -1\right)=0$
The factors of above equation are,
$\left(\mathrm{tan}\theta -2\right)=0\dots \left(1\right)$
$16{\mathrm{sin}}^{2}\theta -1=0\dots \left(2\right)$
Consider the factors.
Add 2 both sides in equation (1).
$\mathrm{tan}\theta =2$
$\mathrm{tan}\theta ={\mathrm{tan}}^{-1}\left(2\right)$
$\theta =1.10714$
Here the angles are in radian.
The tangent has period, $\pi$, so we get all solutions of the equation by adding integer multiples of $\pi$ to these solutions:
${\theta }_{1}=1.10714+k\pi$
The solution obtained for the factor in which sine function involved so we will get the solution in the interval of $\left[0,2\pi \right]$.
Add 1 to both sides in equation (2).
$16{\mathrm{sin}}^{2}\theta =1$
$\mathrm{sin}\theta =±\frac{1}{4}$
$\theta ={\mathrm{sin}}^{-1}\left(±\frac{1}{4}\right)$
Taking positive sign,
$\theta ={\mathrm{sin}}^{-1}\left(\frac{1}{4}\right)$
$=0.252,2.889$
Taking negative sign,
$\theta ={\mathrm{sin}}^{-1}\left(\frac{1}{4}\right)$
$=-0.252,3.393$
The sine has period, $2\pi$, but it is a square function so repeating is done every length of $\pi$. So we get all solutions of the equation by adding integer multiples of $\pi$ to these solutions:
${\theta }_{2}=0.252+2k\pi$
${\theta }_{3}=2.889+2k\pi$
${\theta }_{4}=-0.252+2k\pi$
${\theta }_{5}=3.393+2k\pi$
Therefore, the solutions of the trigonometric equation $\left(\mathrm{tan}\theta -2\right)\left(16{\mathrm{sin}}^{2}\theta -1\right)=0$ are ${\theta }_{1}=1.10714+k\pi ,{\theta }_{2}=0.252+2k\pi ,{\theta }_{3}=2.889+2k\pi ,{\theta }_{4}=-0.252+2k\pi$ and ${\theta }_{5}=3.393+2k\pi$.