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








11 and 12 Grade Math

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