What is the general solution for 5sin(2θ)-7sin(θ)=0 ?

The general solution for 5sin(2θ)-7sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=0.79539…+2pin,θ=2pi-0.79539…+2pin

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5sin(2θ)-7sin(θ)=0

5 sin (2 θ) - 7 sin (θ) = 0

θ = 2 πn, θ = π + 2 πn, θ = 0.79539 \dots + 2 πn, θ = 2 π - 0.79539 \dots + 2 πn

Solution steps

5 sin (2 θ) - 7 sin (θ) = 0

Rewrite using trig identities

- 7 sin (θ) + 10 cos (θ) sin (θ) = 0

Factor

- 7 sin (θ) + 10 cos (θ) sin (θ) : sin (θ) (10 cos (θ) - 7)

sin (θ) (10 cos (θ) - 7) = 0

Solving each part separately

sin (θ) = 0 or 10 cos (θ) - 7 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

10 cos (θ) - 7 = 0 : θ = arccos (\frac{7}{10}) + 2 πn, θ = 2 π - arccos (\frac{7}{10}) + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arccos (\frac{7}{10}) + 2 πn, θ = 2 π - arccos (\frac{7}{10}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = 0.79539 \dots + 2 πn, θ = 2 π - 0.79539 \dots + 2 πn

cos (θ) = \frac{3}{3}

arctan (x + 1) + arctan (x - 1) = arctan (2)

2 sin^{2} (x) - 2 sin (x) = 0

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

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

What is the general solution for 5sin(2θ)-7sin(θ)=0 ?
The general solution for 5sin(2θ)-7sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=0.79539…+2pin,θ=2pi-0.79539…+2pin