# 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