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

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

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sin(4θ)=sin(2θ)

sin (4 θ) = sin (2 θ)

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)

Subtract

sin (2 θ)

from both sides

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

y, x sin (y) = x

cos (2 x) cos (x) + sin (2 x) sin (x) = - 1

2 tan^{2} (x) = - 3 + 3 sec (x)

5 csc^{2} (\frac{θ}{2}) - 16 = 0

sin (3 x) sin (x) = 0, 0 \leq x \leq 2 π

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