The given equation (tan⁡θ−2)(16sin2θ−1)=0

Question

l1koV · Accepted Answer

Approach:
The domain of the trigonometry function of

\tan θ

lies between

[- \infty, \infty]

. Tangent has period

π

, we findd solution in any interval of length

π

.
The domain of the trigonometric function

\sin θ

lies between

[- 1, 1]

. No solution exists beyond this domain. Sine has period

2 π

, we find solution in any interval of length

2 π

. Sine function is positive in first and second quadrant.
Calculation:
The trigonometric equation is given by,

(\tan θ - 2) (16 \sin^{2} θ - 1) = 0

The factors of above equation are,

(\tan θ - 2) = 0 \dots (1)

16 \sin^{2} θ - 1 = 0 \dots (2)

Consider the factors.
Add 2 both sides in equation (1).

\tan θ = 2

\tan θ = \tan^{- 1} (2)

θ = 1.10714

Here the angles are in radian.
The tangent has period,

π

, so we get all solutions of the equation by adding integer multiples of

π

to these solutions:

θ_{1} = 1.10714 + k π

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

[0, 2 π]

.
Add 1 to both sides in equation (2).

16 \sin^{2} θ = 1

\sin θ = \pm \frac{1}{4}

θ = \sin^{- 1} (\pm \frac{1}{4})

Taking positive sign,

θ = \sin^{- 1} (\frac{1}{4})

= 0.252, 2.889

Taking negative sign,

θ = \sin^{- 1} (\frac{1}{4})

= - 0.252, 3.393

The sine has period,

2 π

, but it is a square function so repeating is done every length of

π

. So we get all solutions of the equation by adding integer multiples of

π

to these solutions:

θ_{2} = 0.252 + 2 k π

θ_{3} = 2.889 + 2 k π

θ_{4} = - 0.252 + 2 k π

θ_{5} = 3.393 + 2 k π

Therefore, the solutions of the trigonometric equation

(\tan θ - 2) (16 \sin^{2} θ - 1) = 0

are

θ_{1} = 1.10714 + k π, θ_{2} = 0.252 + 2 k π, θ_{3} = 2.889 + 2 k π, θ_{4} = - 0.252 + 2 k π

and

θ_{5} = 3.393 + 2 k π

.

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