sin θ = sin ∝

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

 Trigonometric Equations





11 and 12 Grade Math

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