What is the general solution for 1-2sin(θ)=0 ?

The general solution for 1-2sin(θ)=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin

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1-2sin(θ)=0

1 - 2 sin (θ) = 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

1 - 2 sin (θ) = 0

Move

1

to the right side

1 - 2 sin (θ) = 0

Subtract

1

from both sides

1 - 2 sin (θ) - 1 = 0 - 1

Simplify

- 2 sin (θ) = - 1

- 2 sin (θ) = - 1

Divide both sides by

- 2

- 2 sin (θ) = - 1

Divide both sides by

- 2

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

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

cos (2 θ) + 3 cos (θ) = 4

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

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

sin (x) = - \frac{1}{6}

sin (θ) = \frac{8}{14}

What is the general solution for 1-2sin(θ)=0 ?
The general solution for 1-2sin(θ)=0 is θ= pi/6+2pin,θ=(5pi)/6+2pin