What is the general solution for sin(t)=-(sqrt(3))/2 ?

The general solution for sin(t)=-(sqrt(3))/2 is t=(4pi)/3+2pin,t=(5pi)/3+2pin

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sin(t)=-(sqrt(3))/2

sin (t) = - \frac{3}{2}

t = \frac{4 π}{3} + 2 πn, t = \frac{5 π}{3} + 2 πn

Degrees

t = 24 0^{\circ} + 36 0^{\circ} n, t = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (t) = - \frac{3}{2}

General solutions for

sin (t) = - \frac{3}{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{4 π}{3} + 2 πn, t = \frac{5 π}{3} + 2 πn

t = \frac{4 π}{3} + 2 πn, t = \frac{5 π}{3} + 2 πn

2 sin (2 θ) - 3 sin (θ) = 0

cos^{2} (x) + cos^{2} (\frac{x}{2}) - \frac{1}{2} = 0

4 cos^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

cos (2 x) + 3 sin (x) = 1

tan (2 θ) - 2 cos (θ) = 0

What is the general solution for sin(t)=-(sqrt(3))/2 ?
The general solution for sin(t)=-(sqrt(3))/2 is t=(4pi)/3+2pin,t=(5pi)/3+2pin