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

The general solution for 5sin(θ)-2=0 is θ=0.41151…+2pin,θ=pi-0.41151…+2pin

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5sin(θ)-2=0

5 sin (θ) - 2 = 0

θ = 0.41151 \dots + 2 πn, θ = π - 0.41151 \dots + 2 πn

Degrees

θ = 23.57817 \dots^{\circ} + 36 0^{\circ} n, θ = 156.42182 \dots^{\circ} + 36 0^{\circ} n

Solution steps

5 sin (θ) - 2 = 0

Move

2

to the right side

5 sin (θ) - 2 = 0

Add

2

to both sides

5 sin (θ) - 2 + 2 = 0 + 2

Simplify

5 sin (θ) = 2

5 sin (θ) = 2

Divide both sides by

5

5 sin (θ) = 2

Divide both sides by

5

\frac{5 sin ( θ )}{5} = \frac{2}{5}

Simplify

sin (θ) = \frac{2}{5}

sin (θ) = \frac{2}{5}

Apply trig inverse properties

sin (θ) = \frac{2}{5}

General solutions for

sin (θ) = \frac{2}{5}

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π - arcsin (a) + 2 πn

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

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

Show solutions in decimal form

θ = 0.41151 \dots + 2 πn, θ = π - 0.41151 \dots + 2 πn

(2 cos (x) - 2) (2 sin (x) - 2) = 0

- 4 cos (5 x) + 1 = 1

- 5 sin (x) = - 2 cos^{2} (x) + 4

6 sin^{2} (θ) - 13 sin (θ) + 7 = 0

sin (x) = 0.1

What is the general solution for 5sin(θ)-2=0 ?
The general solution for 5sin(θ)-2=0 is θ=0.41151…+2pin,θ=pi-0.41151…+2pin