How to find the general solution of the equation tan θ = 0?
Prove that the general solution of tan θ = 0 is θ = nπ, n ∈ Z.
Solution:
According to the figure, by definition, we have,
Tangent function is defined as the ratio of the side perpendicular divided by the adjacent.
Let O be the centre of a unit circle. We know that in unit circle, the length of the circumference is 2π.If we started from A and moves in anticlockwise direction then at the points A, B, A', B' and A, the arc length travelled are 0, \(\frac{π}{2}\), π, \(\frac{3π}{2}\), and 2π.
tan θ = \(\frac{PM}{OM}\)
Now, tan θ = 0
⇒ \(\frac{PM}{OM}\) = 0
⇒ PM = 0.
So when will the tangent be equal to zero?
Clearly, if PM = 0 then the final arm OP of the angle θ coincides with OX or OX'.
Similarly, the final arm OP coincides with OX or OX' when θ = π, 2π, 3π, 4π, ……….. , - π, -2π, -3π, -4π, ……….. i.e. when θ an integral multiples of π i.e., when θ = nπ where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)
Hence, θ = nπ, n ∈ Z is the general solution of the given equation tan θ = 0
1. Find the general solution of the equation tan 2x = 0
Solution:
tan 2x = 0
⇒ 2x = nπ, where, n = 0, ± 1, ± 2, ± 3, ……. [Since, we know that the general solution of the given equation tan θ = 0 is nπ, where, n = 0, ± 1, ± 2, ± 3, ……. ]
⇒ x = \(\frac{nπ}{2}\), where, n = 0, ± 1, ± 2, ± 3, …….
Therefore, the general solution of the trigonometric equation tan 2x = 0 is
x = \(\frac{nπ}{2}\), where, n = 0, ± 1, ± 2, ± 3, …….
2. Find the general solution of the equation tan \(\frac{x}{2}\) = 0
Solution:
tan \(\frac{x}{2}\) = 0
⇒ \(\frac{x}{2}\) = nπ, where, n = 0, ± 1, ± 2, ± 3, ……. [Since, we know that the general solution of the given equation tan θ = 0 is nπ, where, n = 0, ± 1, ± 2, ± 3, ……. ]
⇒ x = 2nπ, where, n = 0, ± 1, ± 2, ± 3, …….
Therefore, the general solution of the trigonometric equation tan \(\frac{x}{2}\) = 0 is
x = 2nπ, where, n = 0, ± 1, ± 2, ± 3, …….
3. What is the general solution of the equation tan x + tan 2x + tan 3x = tan x tan 2x tan 3x?
Solution:
tan x + tan 2x + tan 3x = tan x tan 2x tan 3x
⇒ tan x + tan 2x = - tan 3x + tan x tan 2x tan 3x
⇒ tan x + tan 2x = - tan 3x(1 - tan x tan 2x)
⇒ \(\frac{tan x + tan 2x}{1 - tan x tan 2x}\) = - tan 3x
⇒ tan (x + 2x) = - tan 3x
⇒ tan 3x = - tan 3x
⇒ 2 tan 3x = 0
⇒ tan 3x = 0
⇒ 3x = nπ, where n = 0, ± 1, ± 2, ± 3,…….
x = \(\frac{nπ}{3}\), where n = 0, ± 1, ± 2, ± 3,…….
Therefore, the general solution of the trigonometric equation tan x + tan 2x + tan 3x = tan x tan 2x tan 3x is x = \(\frac{nπ}{3}\), where n = 0, ± 1, ± 2, ± 3,…….
4. Find the general solution of the equation tan \(\frac{3x}{4}\) = 0
Solution:
tan \(\frac{3x}{4}\) = 0
⇒ \(\frac{3x}{4}\) = nπ, where, n = 0, ± 1, ± 2, ± 3, ……. [Since, we know that the general solution of the given equation tan θ = 0 is nπ, where, n = 0, ± 1, ± 2, ± 3, ……. ]
⇒ x = \(\frac{4nπ}{3}\), where, n = 0, ± 1, ± 2, ± 3, …….
Therefore, the general solution of the trigonometric equation tan \(\frac{3x}{4}\) = 0 is x = \(\frac{4nπ}{3}\), where, n = 0, ± 1, ± 2, ± 3, …….
11 and 12 Grade Math
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