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

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

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sin(t)=(-1)/2

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

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

Degrees

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

Solution steps

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

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

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

General solutions for

sin (t) = - \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}

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

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

sin (x + \frac{π}{6}) + cos (x + \frac{π}{3}) = cos (2 x)

24 sin (2 t) - 24 cos (t) = 0

8^{s i n^{2} (x)} = 4^{s i n (x) - \frac{1}{8}}

2 cos (2 x) = \frac{1}{2}

\frac{sin ( 7 3 ^{\circ} )}{34} = \frac{sin ( d )}{29}

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