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

The general solution for 3sin(2θ)=2sin(θ) is θ=2pin,θ=pi+2pin,θ=1.23095…+2pin,θ=2pi-1.23095…+2pin

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3sin(2θ)=2sin(θ)

3 sin (2 θ) = 2 sin (θ)

θ = 2 πn, θ = π + 2 πn, θ = 1.23095 \dots + 2 πn, θ = 2 π - 1.23095 \dots + 2 πn

Solution steps

3 sin (2 θ) = 2 sin (θ)

Subtract

2 sin (θ)

from both sides

3 sin (2 θ) - 2 sin (θ) = 0

Rewrite using trig identities

- 2 sin (θ) + 6 cos (θ) sin (θ) = 0

Factor

- 2 sin (θ) + 6 cos (θ) sin (θ) : 2 sin (θ) (3 cos (θ) - 1)

2 sin (θ) (3 cos (θ) - 1) = 0

Solving each part separately

sin (θ) = 0 or 3 cos (θ) - 1 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

3 cos (θ) - 1 = 0 : θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arccos (\frac{1}{3}) + 2 πn, θ = 2 π - arccos (\frac{1}{3}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = 1.23095 \dots + 2 πn, θ = 2 π - 1.23095 \dots + 2 πn

sin (x) \cdot cos^{2} (x) = sin (x)

sinh (y) = - 2

2 sin (θ) = - 3 + 4 sin (θ)

7^{2} = 6^{2} + 1 1^{2} - 2 \cdot 6 \cdot 11 \cdot cos (D)

- 4 cos^{2} (x) + 3 cos (x) + 1 = 0

What is the general solution for 3sin(2θ)=2sin(θ) ?
The general solution for 3sin(2θ)=2sin(θ) is θ=2pin,θ=pi+2pin,θ=1.23095…+2pin,θ=2pi-1.23095…+2pin