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

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

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sin(4θ)-sin(2θ)=0

sin (4 θ) - sin (2 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 (4 θ) - sin (2 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

tan (x) sin (x) - tan (x) = 0

cos^{2} (θ) - 7 cos (θ) + 12 = 0

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

csc^{2} (θ) - csc (θ) - 20 = 0

sec^{2} (θ) + 7 sec (θ) + 6 = 0

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