What is the general solution for 7sin(2θ)-2sin(θ)=0 ?

The general solution for 7sin(2θ)-2sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=1.42744…+2pin,θ=2pi-1.42744…+2pin

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7sin(2θ)-2sin(θ)=0

7 sin (2 θ) - 2 sin (θ) = 0

θ = 2 πn, θ = π + 2 πn, θ = 1.42744 \dots + 2 πn, θ = 2 π - 1.42744 \dots + 2 πn

Solution steps

7 sin (2 θ) - 2 sin (θ) = 0

Rewrite using trig identities

- 2 sin (θ) + 14 cos (θ) sin (θ) = 0

Factor

- 2 sin (θ) + 14 cos (θ) sin (θ) : 2 sin (θ) (7 cos (θ) - 1)

2 sin (θ) (7 cos (θ) - 1) = 0

Solving each part separately

sin (θ) = 0 or 7 cos (θ) - 1 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

7 cos (θ) - 1 = 0 : θ = arccos (\frac{1}{7}) + 2 πn, θ = 2 π - arccos (\frac{1}{7}) + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arccos (\frac{1}{7}) + 2 πn, θ = 2 π - arccos (\frac{1}{7}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = 1.42744 \dots + 2 πn, θ = 2 π - 1.42744 \dots + 2 πn

cot (2 x + \frac{π}{3}) - 3 = 0, - π < x < π

cos^{4} (x) = \frac{1}{2}

tan (x) = tan (2 x - 30)

2 sin (\frac{x}{2} + \frac{π}{6}) = 1, x < 4 π, 0

16 cos^{2} (x) - 9 = 0

What is the general solution for 7sin(2θ)-2sin(θ)=0 ?
The general solution for 7sin(2θ)-2sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=1.42744…+2pin,θ=2pi-1.42744…+2pin