What is the general solution for 5sin(2θ)-8sin(θ)=0 ?

The general solution for 5sin(2θ)-8sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=0.64350…+2pin,θ=2pi-0.64350…+2pin

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5sin(2θ)-8sin(θ)=0

5 sin (2 θ) - 8 sin (θ) = 0

θ = 2 πn, θ = π + 2 πn, θ = 0.64350 \dots + 2 πn, θ = 2 π - 0.64350 \dots + 2 πn

Solution steps

5 sin (2 θ) - 8 sin (θ) = 0

Rewrite using trig identities

- 8 sin (θ) + 10 cos (θ) sin (θ) = 0

Factor

- 8 sin (θ) + 10 cos (θ) sin (θ) : 2 sin (θ) (5 cos (θ) - 4)

2 sin (θ) (5 cos (θ) - 4) = 0

Solving each part separately

sin (θ) = 0 or 5 cos (θ) - 4 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

5 cos (θ) - 4 = 0 : θ = arccos (\frac{4}{5}) + 2 πn, θ = 2 π - arccos (\frac{4}{5}) + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn, θ = arccos (\frac{4}{5}) + 2 πn, θ = 2 π - arccos (\frac{4}{5}) + 2 πn

Show solutions in decimal form

θ = 2 πn, θ = π + 2 πn, θ = 0.64350 \dots + 2 πn, θ = 2 π - 0.64350 \dots + 2 πn

tan (θ) = \frac{5 3}{5}

f, r = \frac{8}{sec ( f ^{0} )}

2 sin (x) - 2 = 0, 0 \leq x \leq 2 π

(2 sin (x) - 1) (sin (x) + 1) = 0

sin (A) = \frac{3}{4}

What is the general solution for 5sin(2θ)-8sin(θ)=0 ?
The general solution for 5sin(2θ)-8sin(θ)=0 is θ=2pin,θ=pi+2pin,θ=0.64350…+2pin,θ=2pi-0.64350…+2pin