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

The general solution for sin(x)=-(sqrt(3))/2 ,0<= x<= 2pi is x=(4pi)/3 ,x=(5pi)/3

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sin(x)=-(sqrt(3))/2 ,0<= x<= 2pi

sin (x) = - \frac{3}{2}, 0 \leq x \leq 2 π

x = \frac{4 π}{3}, x = \frac{5 π}{3}

Degrees

x = 24 0^{\circ}, x = 30 0^{\circ}

Solution steps

sin (x) = - \frac{3}{2}, 0 \leq x \leq 2 π

General solutions for

sin (x) = - \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}

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

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{4 π}{3}, x = \frac{5 π}{3}

sin^{2} (x) = 1 + cos^{2} (x)

cos^{2} (θ) = 0

cos (7 x) cos (2 x) + sin (7 x) sin (2 x) = \frac{1}{2}

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

8 sin^{2} (θ) - 6 = 0

What is the general solution for sin(x)=-(sqrt(3))/2 ,0<= x<= 2pi ?
The general solution for sin(x)=-(sqrt(3))/2 ,0<= x<= 2pi is x=(4pi)/3 ,x=(5pi)/3