What is the general solution for sin(θ)-sin(2θ)=0 ?

The general solution for sin(θ)-sin(2θ)=0 is θ= pi/3+(4pin)/3 ,θ=pi+(4pin)/3 ,θ=4pin,θ=2pi+4pin

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sin(θ)-sin(2θ)=0

sin (θ) - sin (2 θ) = 0

θ = \frac{π}{3} + \frac{4 πn}{3}, θ = π + \frac{4 πn}{3}, θ = 4 πn, θ = 2 π + 4 πn

Solution steps

sin (θ) - sin (2 θ) = 0

Rewrite using trig identities

- 2 cos (\frac{3 θ}{2}) sin (\frac{θ}{2}) = 0

Solving each part separately

cos (\frac{3 θ}{2}) = 0 or sin (\frac{θ}{2}) = 0

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

sin (\frac{θ}{2}) = 0 : θ = 4 πn, θ = 2 π + 4 πn

Combine all the solutions

θ = \frac{π}{3} + \frac{4 πn}{3}, θ = π + \frac{4 πn}{3}, θ = 4 πn, θ = 2 π + 4 πn

16 cos (8 x) - 2 = 6

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

4 sin^{2} (θ) - 4 sin (θ) + 1 = 0

2 cos (θ) - 3 = 5 cos (θ) - 5

tan (θ) = \frac{4}{5}

What is the general solution for sin(θ)-sin(2θ)=0 ?
The general solution for sin(θ)-sin(2θ)=0 is θ= pi/3+(4pin)/3 ,θ=pi+(4pin)/3 ,θ=4pin,θ=2pi+4pin