What is the general solution for 2sin(x)=-1,0<= x<2pi ?

The general solution for 2sin(x)=-1,0<= x<2pi is x=(7pi)/6 ,x=(11pi)/6

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2sin(x)=-1,0<= x<2pi

2 sin (x) = - 1, 0 \leq x < 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

Solution steps

2 sin (x) = - 1, 0 \leq x < 2 π

Divide both sides by

2

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

General solutions for

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

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

Solutions for the range

0 \leq x < 2 π

x = \frac{7 π}{6}, x = \frac{11 π}{6}

2 cos^{2} (x) = 3 - 3 sin (x)

sec (3 x) - 4 = 0

6 sin^{2} (θ) + 5 sin (θ) = - 2 sin (θ) - 1

t, x = 5 cos (t)

tan (θ) = \frac{π}{2}

What is the general solution for 2sin(x)=-1,0<= x<2pi ?
The general solution for 2sin(x)=-1,0<= x<2pi is x=(7pi)/6 ,x=(11pi)/6