cos θ = cos ∝

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

Prove that the general solution of cos θ = cos ∝ is given by θ = 2nπ ± ∝, n ∈ Z.

Solution:

We have,

cos θ = cos ∝

⇒ cos θ - cos ∝ = 0 

⇒ 2 sin \(\frac{(θ   +  ∝)}{2}\) sin \(\frac{(θ   -  ∝)}{2}\) = 0

Therefore, either, sin \(\frac{(θ   +  ∝)}{2}\) = 0 or, sin \(\frac{(θ   -  ∝)}{2}\) = 0

Now, from sin \(\frac{(θ   +  ∝)}{2}\) = 0 we get, \(\frac{(θ   +  ∝)}{2}\) = nπ, n ∈ Z

⇒ θ = 2nπ - ∝, n ∈ Z i.e., (any even multiple of π) - ∝ …………………….(i)

And from sin \(\frac{(θ   -  ∝)}{2}\) = 0 we get,

\(\frac{(θ   -  ∝)}{2}\) = nπ, n ∈ Z                  

⇒ θ = 2nπ + ∝, m ∈ Z i.e., (any even multiple of π) + ∝ …………………….(ii)

Now combining the solutions (i) and (ii) we get,

θ = 2nπ ± ∝, where n ∈ Z.

Hence, the general solution of cos θ = cos ∝ is θ = 2nπ ± , where n ∈ Z.


Note: The equation sec θ = sec ∝ is equivalent to cos θ = cos ∝ (since, sec θ = \(\frac{1}{cos  θ}\) and sec ∝ = \(\frac{1}{cos  ∝}\)). Thus, sec θ = sec ∝ and cos θ = cos ∝ have the same general solution.

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


1. Find the general values of θ if cos θ = - \(\frac{√3}{2}\).

Solution:

cos θ = - \(\frac{√3}{2}\)

⇒ cos θ = - cos \(\frac{π}{6}\)

⇒ cos θ = cos (π - \(\frac{π}{6}\))

⇒ cos θ = cos \(\frac{5π}{6}\)

⇒ θ = 2nπ ± \(\frac{5π}{6}\), where n ∈ Z (i.e., n = 0, ± 1, ± 2, ± 3,…….)


2. Find the general values of θ if cos θ = \(\frac{1}{2}\)

Solution:

cos θ = \(\frac{1}{2}\)

cos θ = cos \(\frac{π}{3}\)

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

Therefore the general solution of cos θ = \(\frac{1}{2}\) is θ = 2nπ ± \(\frac{π}{3}\), where, n = 0, ± 1, ± 2, ± 3, ± 4 .....

3. Solve for x if 0 ≤ x ≤ \(\frac{π}{2}\) sin x + sin 5x = sin 3x

Solution:

sin x + sin 5x = sin 3x

⇒ sin 5x + sin x = sin 3x

⇒ 2 sin \(\frac{5x + x}{2}\) cos \(\frac{5x + x}{2}\) = sin 3x

⇒ 2 sin 3x cos 2x = sin 3x

⇒ 2 sin 3x cos 2x - sin 3x = 0

⇒ sin 3x (2 cos 2x - 1) = 0

Therefore, either sin 3x = 0 or 2 cos 2x – 1 = 0

Now, from sin 3x = 0 we get,

3x = nπ   

⇒ x = \(\frac{nπ}{3}\) …………..(1)

similarly, from 2 cos 2x - 1 = 0 we get,

⇒ cos 2x = \(\frac{1}{2}\)

⇒ cos 2x = cos \(\frac{π}{3}\)

Therefore, 2x = 2nπ ± \(\frac{π}{3}\)

⇒ x = nπ ± \(\frac{π}{6}\) …………..(2)

Now, putting n = 0 in (1) we get, x = 0            

Now, putting n = 1 in (1) we get, x = \(\frac{π}{3}\)       

Now, putting n = 0 in (2) we get, x = ± \(\frac{π}{6}\)      

Therefore, the required solutions of the given equation in 0 ≤ x ≤ π/2 are:

x = 0, \(\frac{π}{3}\), \(\frac{π}{6}\).

 Trigonometric Equations












11 and 12 Grade Math

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