# tan θ = tan ∝

How to find the general solution of an equation of the form tan θ = tan ∝?

Prove that the general solution of tan θ = tan ∝ is given by θ = nπ +∝, n ∈ Z.

Solution:

We have,

tan θ = tan ∝

⇒ sin θ/cos θ - sin ∝/cos ∝ = 0

⇒ (sin θ cos ∝ - cos θ sin ∝)/cos θ cos ∝ = 0

⇒ sin (θ - ∝)/cos θ cos ∝ = 0

⇒ sin (θ - ∝) = 0

⇒ sin (θ - ∝) = 0

⇒ (θ - ∝) = nπ, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….), [Since we know that the θ = nπ, n ∈ Z is the general solution of the given equation sin θ = 0]

⇒ θ = nπ + ∝, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Hence, the general solution of tan θ = tan ∝ is θ = nπ + , where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Note: The equation cot θ = cot ∝ is equivalent to tan θ = tan ∝ (since, cot θ = 1/tan θ and cot ∝ = 1/tan ∝). Thus, cot θ = cot ∝ and tan θ = tan ∝ have the same general solution.

Hence, the general solution of cot θ = cot ∝ is θ = nπ + , where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

1. Solve the trigonometric equation tan θ = $$\frac{1}{√3}$$

Solution:

tan θ = $$\frac{1}{√3}$$

⇒ tan θ = tan $$\frac{π}{6}$$

⇒ θ = nπ + $$\frac{π}{6}$$, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….), [Since, we know that the general solution of tan θ = tan ∝ is θ = nπ + ∝, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)]

2. What is the general solution of the trigonometric equation tan x + tan 2x + tan x tan 2x = 1?

Solution:

tan x + tan 2x + tan x tan 2x = 1

tan x + tan 2x = 1 - tan x tan 2x

$$\frac{tan x + tan 2x}{1 - tan x tan 2x}$$ = 1

tan 3x = 1

tan 3x = tan $$\frac{π}{4}$$

3x = nπ + $$\frac{π}{4}$$, where n = 0, ± 1, ± 2, ± 3,…….

x = $$\frac{nπ}{3}$$ + $$\frac{π}{12}$$, where n = 0, ± 1, ± 2, ± 3,…….

Therefore, the general solution of the trigonometric equation tan x + tan 2x + tan x tan 2x = 1 is x = $$\frac{nπ}{3}$$ + $$\frac{π}{12}$$, where n = 0, ± 1, ± 2, ± 3,…….

3. Solve the trigonometric equation tan 2θ = √3

Solution:

tan 2θ = √3

⇒ tan 2θ = tan $$\frac{π}{3}$$

⇒ 2θ = nπ + $$\frac{π}{3}$$, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….), [Since, we know that the general solution of tan θ = tan ∝ is θ = nπ + ∝, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)]

⇒ θ = $$\frac{nπ}{2}$$ + $$\frac{π}{6}$$, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Hence, the general solution of tan 2θ = √3 is θ = $$\frac{nπ}{2}$$ + $$\frac{π}{6}$$, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

4. Find the general solution of the trigonometric equation 2 tan x - cot x + 1 = 0

Solution:

2 tan x - cot x + 1 = 0

⇒ 2 tan x - $$\frac{1}{tan x }$$ + 1 = 0

⇒ 2 tan$$^{2}$$ x + tan x - 1 = 0

⇒ 2 tan$$^{2}$$ x + 2tan x - tan x - 1 = 0

⇒ 2 tan x(tan x + 1) - 1(tan x + 1) = 0

⇒ (tan x + 1)(2 tan x - 1) = 0

⇒ either tan x + 1 = or, 2 tan x - 1 = 0

⇒ tan x = -1 or, tan x  = $$\frac{1}{2}$$

⇒ tan x = ($$\frac{-π}{4}$$) or, tan x  = tan α, where tan α = $$\frac{1}{2}$$

⇒ x = nπ + ($$\frac{-π}{4}$$) or, x = mπ + α, where tan α = $$\frac{1}{2}$$ and m = 0, ± 1, ± 2, ± 3,…….

⇒ x = nπ - ($$\frac{π}{4}$$) or, x = mπ + α, where tan α = $$\frac{1}{2}$$ and m = 0, ± 1, ± 2, ± 3,…….

Therefore the solution of the trigonometric equation 2 tan x - cot x + 1 = 0 are x = nπ - ($$\frac{π}{4}$$)  and x = mπ + α, where tan α = $$\frac{1}{2}$$ and m = 0, ± 1, ± 2, ± 3,…….

5. Solve the trigonometric equation tan 3θ  + 1 = 0

Solution:

tan 3θ  + 1 = 0

tan 3θ  = - 1

⇒ tan 3θ = tan (-$$\frac{π}{4}$$)

⇒ 3θ = nπ + (-$$\frac{π}{4}$$), where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….), [Since, we know that the general solution of tan θ = tan ∝ is θ = nπ + ∝, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)]

⇒ θ = $$\frac{nπ}{3}$$ - $$\frac{π}{12}$$, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Hence, the general solution of tan 3θ  + 1 = 0 is θ = $$\frac{nπ}{3}$$ - $$\frac{π}{12}$$, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Trigonometric Equations

Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.