Question

Solve the following equations:
sin θ + sin 3θ + sin 5θ = 0

Answer 1

sin θ + sin 3θ + sin 5θ = 0

`2 sin ((5theta + theta)/2) * cos ((5theta - theta)/2) + sin 3theta` = 0

`2sin ((6theta)/2) * cos ((4theta)/2) + sin 3theta` = 0

2 sin 3θ . cos 2θ + sin 3θ = 0

sin 3θ (2 cos 2θ + 1) = θ

sin 3θ = 0 or 2 cos 2θ + 1 = θ

sin 3θ = 0 or cos 2θ = `- 1/2`

To find the general solution of sin 3θ = 0

The general solution is

3θ = nπ, n ∈ Z

θ = `("n"pi)/3`, n ∈ Z

To find the general solution of cos 2θ = ` - 1/2`

cos 2θ = ` - 1/2`

cos 2θ = `cos (pi - pi/3)`

cos 2θ = `cos ((3pi - pi)/3)`

cos 2θ = `cos  ((2pi)/3)`

The general solution is

2θ = `2"n"pi +- (2pi)/3`, n ∈ Z

θ = `"n"pi +-  pi/3`, n ∈ Z

∴ The required solutions are

θ = `(:"n"pi)/3`, n ∈ Z

or

θ = `"n"pi +-  pi/3`, n ∈ Z