What is the general solution for 2sin(2θ)-3sin(θ)=0 ?

The general solution for 2sin(2θ)-3sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=0.72273…+2pin,θ=2pi-0.72273…+2pin

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2sin(2θ)-3sin(θ)=0

2 sin (2 θ) - 3 sin (θ) = 0

θ = 2 πn, θ = π + 2 πn, θ = 0.72273 \dots + 2 πn, θ = 2 π - 0.72273 \dots + 2 πn

Solution steps

2 sin (2 θ) - 3 sin (θ) = 0

Rewrite using trig identities

- 3 sin (θ) + 4 cos (θ) sin (θ) = 0

Factor

- 3 sin (θ) + 4 cos (θ) sin (θ) : sin (θ) (4 cos (θ) - 3)

sin (θ) (4 cos (θ) - 3) = 0

Solving each part separately

sin (θ) = 0 or 4 cos (θ) - 3 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

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

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arccos (\frac{3}{4}) + 2 πn, θ = 2 π - arccos (\frac{3}{4}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = 0.72273 \dots + 2 πn, θ = 2 π - 0.72273 \dots + 2 πn

cos^{2} (x) + cos^{2} (\frac{x}{2}) - \frac{1}{2} = 0

4 cos^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

cos (2 x) + 3 sin (x) = 1

tan (2 θ) - 2 cos (θ) = 0

tan (θ) = \frac{12}{5}, π < θ < \frac{3 π}{2}

What is the general solution for 2sin(2θ)-3sin(θ)=0 ?
The general solution for 2sin(2θ)-3sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=0.72273…+2pin,θ=2pi-0.72273…+2pin