What is the general solution for sin(θ)=(sqrt(3))/2 ,0<= θ<= 4pi ?

The general solution for sin(θ)=(sqrt(3))/2 ,0<= θ<= 4pi is θ= pi/3 ,θ=(2pi)/3 ,θ=(7pi)/3 ,θ=(8pi)/3

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sin(θ)=(sqrt(3))/2 ,0<= θ<= 4pi

sin (θ) = \frac{3}{2}, 0 \leq θ \leq 4 π

θ = \frac{π}{3}, θ = \frac{2 π}{3}, θ = \frac{7 π}{3}, θ = \frac{8 π}{3}

Degrees

θ = 6 0^{\circ}, θ = 12 0^{\circ}, θ = 42 0^{\circ}, θ = 48 0^{\circ}

Solution steps

sin (θ) = \frac{3}{2}, 0 \leq θ \leq 4 π

General solutions for

sin (θ) = \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}

θ = \frac{π}{3} + 2 πn, θ = \frac{2 π}{3} + 2 πn

θ = \frac{π}{3} + 2 πn, θ = \frac{2 π}{3} + 2 πn

Solutions for the range

0 \leq θ \leq 4 π

θ = \frac{π}{3}, θ = \frac{2 π}{3}, θ = \frac{7 π}{3}, θ = \frac{8 π}{3}

sin (2 x) = \frac{3}{2}, 0^{\circ} \leq x \leq 36 0^{\circ}

3 sin^{2} (x) + 8 cos^{2} (x) = 6

tan (θ) = \frac{2}{8}

4 tan^{2} (x) + 7 tan (x) - 4 = 0, cot (2 x)

tan^{3} (x) = tan^{2} (x), 0 \leq x \leq 2 π

What is the general solution for sin(θ)=(sqrt(3))/2 ,0<= θ<= 4pi ?
The general solution for sin(θ)=(sqrt(3))/2 ,0<= θ<= 4pi is θ= pi/3 ,θ=(2pi)/3 ,θ=(7pi)/3 ,θ=(8pi)/3