What is the general solution for sin(2x-5 pi/6)=-1/2 ,0<= x<= 2pi ?

The general solution for sin(2x-5 pi/6)=-1/2 ,0<= x<= 2pi is x=pi,x=(4pi)/3 ,x=2pi,x=0,x= pi/3

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sin(2x-5 pi/6)=-1/2 ,0<= x<= 2pi

sin (2 x - 5 \frac{π}{6}) = - \frac{1}{2}, 0 \leq x \leq 2 π

x = π, x = \frac{4 π}{3}, x = 2 π, x = 0, x = \frac{π}{3}

Solution steps

sin (2 x - 5 \cdot \frac{π}{6}) = - \frac{1}{2}, 0 \leq x \leq 2 π

General solutions for

sin (2 x - 5 \frac{π}{6}) = - \frac{1}{2}

2 x - 5 \cdot \frac{π}{6} = \frac{7 π}{6} + 2 πn, 2 x - 5 \cdot \frac{π}{6} = \frac{11 π}{6} + 2 πn

Solve

2 x - 5 \frac{π}{6} = \frac{7 π}{6} + 2 πn : x = πn + π

Solve

2 x - 5 \frac{π}{6} = \frac{11 π}{6} + 2 πn : x = πn + \frac{4 π}{3}

x = πn + π, x = πn + \frac{4 π}{3}

Solutions for the range

0 \leq x \leq 2 π

x = π, x = \frac{4 π}{3}, x = 2 π, x = 0, x = \frac{π}{3}

sin (x) = 1, 0 \leq x \leq 2 π

6 csc (θ) - 6 = 0

9 sin (x) tan (x) + 4 tan (x) = 0

x, f [g] = cos (\frac{1}{x ^{2}})

2 cos (\frac{π}{5} θ) = 3

What is the general solution for sin(2x-5 pi/6)=-1/2 ,0<= x<= 2pi ?
The general solution for sin(2x-5 pi/6)=-1/2 ,0<= x<= 2pi is x=pi,x=(4pi)/3 ,x=2pi,x=0,x= pi/3