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

• General solution of the equation sin x = ½
• General solution of the equation cos x = 1/√2
• General solution of the equation tan x = √3
• General Solution of the Equation sin θ = 0
• General Solution of the Equation cos θ = 0
• General Solution of the Equation tan θ = 0
• General Solution of the Equation sin θ = sin ∝
  
• General Solution of the Equation sin θ = 1
• General Solution of the Equation sin θ = -1
• General Solution of the Equation cos θ = cos ∝
• General Solution of the Equation cos θ = 1
• General Solution of the Equation cos θ = -1
• General Solution of the Equation tan θ = tan ∝
• General Solution of a cos θ + b sin θ = c
• Trigonometric Equation Formula
• Trigonometric Equation using Formula
• General solution of Trigonometric Equation
• Problems on Trigonometric Equation

11 and 12 Grade Math

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