What is the general solution for sin(3θ)-sin(5θ)=sin(θ),0<= θ<= pi/2 ?

The general solution for sin(3θ)-sin(5θ)=sin(θ),0<= θ<= pi/2 is θ=0,θ= pi/6 ,θ= pi/3

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sin(3θ)-sin(5θ)=sin(θ),0<= θ<= pi/2

sin (3 θ) - sin (5 θ) = sin (θ), 0 \leq θ \leq \frac{π}{2}

θ = 0, θ = \frac{π}{6}, θ = \frac{π}{3}

Solution steps

sin (3 θ) - sin (5 θ) = sin (θ), 0 \leq θ \leq \frac{π}{2}

Subtract

sin (θ)

from both sides

sin (3 θ) - sin (5 θ) - sin (θ) = 0

Rewrite using trig identities

- sin (θ) - 2 cos (4 θ) sin (θ) = 0

Factor

- sin (θ) - 2 cos (4 θ) sin (θ) : - sin (θ) (2 cos (4 θ) + 1)

- sin (θ) (2 cos (4 θ) + 1) = 0

Solving each part separately

sin (θ) = 0 or 2 cos (4 θ) + 1 = 0

sin (θ) = 0, 0 \leq θ \leq \frac{π}{2} : θ = 0

2 cos (4 θ) + 1 = 0, 0 \leq θ \leq \frac{π}{2} : θ = \frac{π}{6}, θ = \frac{π}{3}

Combine all the solutions

θ = 0, θ = \frac{π}{6}, θ = \frac{π}{3}

4 sin (2 θ) - 1 = 0

\frac{3}{2} = sin (x)

sec (w) csc (w) = 2 sec (w)

- 2 sin (x) - cos (2 x) \cdot 2 = 0

- 2 cos (B) + 4 = cos (B) + 6

What is the general solution for sin(3θ)-sin(5θ)=sin(θ),0<= θ<= pi/2 ?
The general solution for sin(3θ)-sin(5θ)=sin(θ),0<= θ<= pi/2 is θ=0,θ= pi/6 ,θ= pi/3