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, \(\frac{π}{3}\), \(\frac{π}{6}\).
11 and 12 Grade Math
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