What is the general solution for sin(2θ)-sin(4θ)=0 ?

The general solution for sin(2θ)-sin(4θ)=0 is θ= pi/6+(2pin)/3 ,θ= pi/2+(2pin)/3 ,θ=2pin,θ=pi+2pin

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sin(2θ)-sin(4θ)=0

sin (2 θ) - sin (4 θ) = 0

θ = \frac{π}{6} + \frac{2 πn}{3}, θ = \frac{π}{2} + \frac{2 πn}{3}, θ = 2 πn, θ = π + 2 πn

Solution steps

sin (2 θ) - sin (4 θ) = 0

Rewrite using trig identities

- 2 cos (3 θ) sin (θ) = 0

Solving each part separately

cos (3 θ) = 0 or sin (θ) = 0

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

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

Combine all the solutions

θ = \frac{π}{6} + \frac{2 πn}{3}, θ = \frac{π}{2} + \frac{2 πn}{3}, θ = 2 πn, θ = π + 2 πn

2 cos^{2} (x) + 7 cos (x) + 3 = 0

3 cos (x) + 2 = - cos (x)

2 cos (x) + 2 sin (x) tan (x) = 4

3 sin (x) tan (x) = 8

tan^{2} (x) - 4 = 0

What is the general solution for sin(2θ)-sin(4θ)=0 ?
The general solution for sin(2θ)-sin(4θ)=0 is θ= pi/6+(2pin)/3 ,θ= pi/2+(2pin)/3 ,θ=2pin,θ=pi+2pin