# 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

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