Loading [MathJax]/jax/output/HTML-CSS/jax.js

Subscribe to our ▶️ YouTube channel 🔴 for the latest videos, updates, and tips.

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 = 13

Solution:

tan θ = 13

⇒ tan θ = tan π6

⇒ θ = nπ + π6, 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

tanx+tan2x1tanxtan2x = 1

tan 3x = 1

tan 3x = tan π4

3x = nπ + π4, where n = 0, ± 1, ± 2, ± 3,…….

x = nπ3 + π12, 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 = nπ3 + π12, where n = 0, ± 1, ± 2, ± 3,…….


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

Solution:

tan 2θ = √3

⇒ tan 2θ = tan π3

⇒ 2θ = nπ + π3where 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 = nπ2 + π6where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Hence, the general solution of tan 2θ = √3 is θ = nπ2 + π6, 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 - 1tanx + 1 = 0

⇒ 2 tan2 x + tan x - 1 = 0

⇒ 2 tan2 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  = 12

⇒ tan x = (π4) or, tan x  = tan α, where tan α = 12

⇒ x = nπ + (π4) or, x = mπ + α, where tan α = 12 and m = 0, ± 1, ± 2, ± 3,…….

⇒ x = nπ - (π4) or, x = mπ + α, where tan α = 12 and m = 0, ± 1, ± 2, ± 3,…….

Therefore the solution of the trigonometric equation 2 tan x - cot x + 1 = 0 are x = nπ - (π4)  and x = mπ + α, where tan α = 12 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 (-π4)

⇒ 3θ = nπ + (-π4), 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 = nπ3 - π12where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

Hence, the general solution of tan 3θ  + 1 = 0 is θ = nπ3 - π12, where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)

 Trigonometric Equations








11 and 12 Grade Math

From tan θ = tan ∝ to HOME PAGE




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.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.