How to find the general solution of an equation of the form sin θ = sin ∝?
Prove that the general solution of sin θ = sin ∝ is given by θ = nπ + (1)\(^{n}\) ∝, n ∈ Z.
Solution:
We have,
sin θ = sin ∝
⇒ sin θ  sin ∝ = 0
⇒ 2 cos \(\frac{θ + ∝}{2}\) sin \(\frac{θ  ∝}{2}\) = 0
Therefore either cos \(\frac{θ + ∝}{2}\) = 0 or, sin \(\frac{θ  ∝}{2}\) = 0
Now, from cos \(\frac{θ + ∝}{2}\) = 0 we
get, \(\frac{θ + ∝}{2}\) = (2m + 1)\(\frac{π}{2}\), m ∈ Z
⇒ θ = (2m + 1)π  ∝, m ∈ Z i.e., (any odd multiple of π)  ∝ ……………….(i)
And from sin \(\frac{θ  ∝}{2}\) = 0 we get,
\(\frac{θ  ∝}{2}\) = mπ, m ∈ Z
⇒ θ = 2mπ + ∝, m ∈ Z i.e., (any even multiple of π) + ∝ …………………….(ii)
Now combining the solutions (i) and (ii) we get,
θ = nπ + (1)\(^{n}\) ∝, where n ∈ Z.
Hence, the general solution of sin θ = sin ∝ is θ = nπ + (1)\(^{n}\) ∝, where n ∈ Z.
Note: The equation csc θ = csc ∝ is equivalent to sin θ = sin ∝ (since, csc θ = \(\frac{1}{sin θ}\) and csc ∝ = \(\frac{1}{sin ∝}\)). Thus, csc θ = csc ∝ and sin θ = sin ∝ have the same general solution.
Hence, the general solution of csc θ = csc ∝ is θ = nπ + (1)\(^{n}\) ∝, where n ∈ Z.
1. Find the general values of x which satisfy the equation sin 2x = \(\frac{1}{2}\)
solution:
sin 2x = \(\frac{1}{2}\)
sin 2x =  sin \(\frac{π}{6}\)
⇒ sin 2x = sin (π + \(\frac{π}{6}\))
⇒ sin 2x = sin \(\frac{7π}{6}\)
⇒ 2x = nπ + (1)\(^{n}\) \(\frac{7π}{6}\), n ∈ Z
⇒ x = \(\frac{nπ}{2}\) + (1)\(^{n}\) \(\frac{7π}{12}\), n ∈ Z
Therefore the general solution of sin 2x = \(\frac{1}{2}\) is x = \(\frac{nπ}{2}\) + (1)\(^{n}\) \(\frac{7π}{12}\), n ∈ Z
2. Find the general solution of the trigonometric equation sin 3θ = \(\frac{√3}{2}\).
Solution:
sin 3θ = \(\frac{√3}{2}\)
⇒ sin 3θ = sin \(\frac{π}{3}\)
⇒ 3θ = = nπ + (1)\(^{n}\) \(\frac{π}{3}\), where, n = 0, ± 1, ± 2, ± 3, ± 4 .....
⇒ θ = \(\frac{nπ}{3}\) + (1)\(^{n}\) \(\frac{π}{9}\),where, n = 0, ± 1, ± 2, ± 3, ± 4 .....
Therefore the general solution of sin 3θ = \(\frac{√3}{2}\) is θ = \(\frac{nπ}{3}\) + (1)\(^{n}\) \(\frac{π}{9}\), where, n = 0, ± 1, ± 2, ± 3, ± 4 .....
3. Find the general solution of the equation csc θ = 2
Solution:
csc θ = 2
⇒ sin θ = \(\frac{1}{2}\)
⇒ sin θ = sin \(\frac{π}{6}\)
⇒ θ = nπ + (1)\(^{n}\) \(\frac{π}{6}\), where, n ∈ Z, [Since, we know that the general solution of the equation sin θ = sin ∝ is θ = 2nπ + (1)\(^{n}\) ∝, where n = 0, ± 1, ± 2, ± 3, ……. ]
Therefore the general solution of csc θ = 2 is θ = nπ + (1)\(^{n}\) \(\frac{π}{6}\), where, n ∈ Z
4. Find the general solution of the trigonometric equation sin\(^{2}\) θ = \(\frac{3}{4}\).
Solution:
sin\(^{2}\) θ = \(\frac{3}{4}\).
⇒ sin θ = ± \(\frac{√3}{2}\)
⇒ sin θ = sin (± \(\frac{π}{3}\))
⇒ θ = nπ + (1)\(^{n}\) ∙ (±\(\frac{π}{3}\)), where, n ∈ Z
⇒ θ = nπ ±\(\frac{π}{3}\), where, n ∈ Z
Therefore the general solution of sin\(^{2}\) θ = \(\frac{3}{4}\) is θ = nπ ±\(\frac{π}{3}\), where, n ∈ Z
11 and 12 Grade Math
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