What is the general solution for 4sin(θ)-2=0 ?

The general solution for 4sin(θ)-2=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin

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4sin(θ)-2=0

4 sin (θ) - 2 = 0

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Degrees

θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

4 sin (θ) - 2 = 0

Move

2

to the right side

4 sin (θ) - 2 = 0

Add

2

to both sides

4 sin (θ) - 2 + 2 = 0 + 2

Simplify

4 sin (θ) = 2

4 sin (θ) = 2

Divide both sides by

4

4 sin (θ) = 2

Divide both sides by

4

\frac{4 sin ( θ )}{4} = \frac{2}{4}

Simplify

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \frac{1}{2}

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

sin (θ) = - \frac{5}{13}

2 sin^{2} (x) + 5 sin (x) - 3 = 0

- \frac{1}{2} = sin (x)

2 sin (θ) + 4 = 5

2 cos (x) - 3 tan (x) = 0

What is the general solution for 4sin(θ)-2=0 ?
The general solution for 4sin(θ)-2=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin