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

Elleanor Mckenzie
2021-08-11
Answered

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

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l1koV

Answered 2021-08-12
Author has **100** answers

Approach:

The domain of the trigonometry function of$\mathrm{tan}\theta$ lies between $[-\mathrm{\infty},\mathrm{\infty}]$ . Tangent has period $\pi$ , we findd solution in any interval of length $\pi$ .

The domain of the trigonometric function$\mathrm{sin}\theta$ lies between $[-1,1]$ . 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,

$(\mathrm{tan}\theta -2)(16{\mathrm{sin}}^{2}\theta -1)=0$

The factors of above equation are,

$(\mathrm{tan}\theta -2)=0\dots \left(1\right)$

$16{\mathrm{sin}}^{2}\theta -1=0\dots \left(2\right)$

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$[0,2\pi ]$ .

Add 1 to both sides in equation (2).

$16{\mathrm{sin}}^{2}\theta =1$

$\mathrm{sin}\theta =\pm \frac{1}{4}$

$\theta ={\mathrm{sin}}^{-1}(\pm \frac{1}{4})$

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$(\mathrm{tan}\theta -2)(16{\mathrm{sin}}^{2}\theta -1)=0$ 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 ${\theta}_{5}=3.393+2k\pi$ .

The domain of the trigonometry function of

The domain of the trigonometric function

Calculation:

The trigonometric equation is given by,

The factors of above equation are,

Consider the factors.

Add 2 both sides in equation (1).

Here the angles are in radian.

The tangent has period,

The solution obtained for the factor in which sine function involved so we will get the solution in the interval of

Add 1 to both sides in equation (2).

Taking positive sign,

Taking negative sign,

The sine has period,

Therefore, the solutions of the trigonometric equation

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