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

Elleanor Mckenzie 2021-08-11 Answered
The given equation (tanθ2)(16sin2θ1)=0
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Expert Answer

l1koV
Answered 2021-08-12 Author has 100 answers
Approach:
The domain of the trigonometry function of tanθ lies between [,]. Tangent has period π, we findd solution in any interval of length π.
The domain of the trigonometric function sinθ lies between [1,1]. No solution exists beyond this domain. Sine has period 2π, we find solution in any interval of length 2π. Sine function is positive in first and second quadrant.
Calculation:
The trigonometric equation is given by,
(tanθ2)(16sin2θ1)=0
The factors of above equation are,
(tanθ2)=0(1)
16sin2θ1=0(2)
Consider the factors.
Add 2 both sides in equation (1).
tanθ=2
tanθ=tan1(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:
θ1=1.10714+kπ
The solution obtained for the factor in which sine function involved so we will get the solution in the interval of [0,2π].
Add 1 to both sides in equation (2).
16sin2θ=1
sinθ=±14
θ=sin1(±14)
Taking positive sign,
θ=sin1(14)
=0.252,2.889
Taking negative sign,
θ=sin1(14)
=0.252,3.393
The sine has period, 2π, 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:
θ2=0.252+2kπ
θ3=2.889+2kπ
θ4=0.252+2kπ
θ5=3.393+2kπ
Therefore, the solutions of the trigonometric equation (tanθ2)(16sin2θ1)=0 are θ1=1.10714+kπ,θ2=0.252+2kπ,θ3=2.889+2kπ,θ4=0.252+2kπ and θ5=3.393+2kπ.
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