# 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.

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

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