cos θ = cos ∝

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, $\frac{π}{3}$, $\frac{π}{6}$.

● 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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