tan θ = 0

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.

tan θ = 0tan θ = 0

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, …….





● Trigonometric Equations






11 and 12 Grade Math 

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