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.

tan 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

• General solution of the equation sin x = ½
• General solution of the equation cos x = 1/√2
• General solution of the equation tan x = √3
• General Solution of the Equation sin θ = 0
• General Solution of the Equation cos θ = 0
• General Solution of the Equation tan θ = 0
• General Solution of the Equation sin θ = sin ∝
  
• General Solution of the Equation sin θ = 1
• General Solution of the Equation sin θ = -1
• General Solution of the Equation cos θ = cos ∝
• General Solution of the Equation cos θ = 1
• General Solution of the Equation cos θ = -1
• General Solution of the Equation tan θ = tan ∝
• General Solution of a cos θ + b sin θ = c
• Trigonometric Equation Formula
• Trigonometric Equation using Formula
• General solution of Trigonometric Equation
• Problems on Trigonometric Equation

11 and 12 Grade Math 

From tan θ = 0 to HOME PAGE

11 and 12 Grade Math

From tan θ = 0 to HOME PAGE