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

The general solution for -2sin(θ)=1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin

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-2sin(θ)=1

- 2 sin (θ) = 1

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Degrees

θ = 21 0^{\circ} + 36 0^{\circ} n, θ = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

- 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{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

cos (θ) = - \frac{7}{3}

0 = - 6 sin (x) + 6 cos (2 x)

tan (2 x + 1) = \frac{1}{2}

cos (B) = \frac{8}{12}

sin (t) = \frac{1}{2}, 0 \leq t < \frac{π}{2}

What is the general solution for -2sin(θ)=1 ?
The general solution for -2sin(θ)=1 is θ=(7pi)/6+2pin,θ=(11pi)/6+2pin