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

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

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cos(2x-pi/6)=-1/2 , pi/2 <= x<pi

cos (2 x - \frac{π}{6}) = - \frac{1}{2}, \frac{π}{2} \leq x < π

x = \frac{3 π}{4}

Solution steps

cos (2 x - \frac{π}{6}) = - \frac{1}{2}, \frac{π}{2} \leq x < π

General solutions for

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

2 x - \frac{π}{6} = \frac{2 π}{3} + 2 πn, 2 x - \frac{π}{6} = \frac{4 π}{3} + 2 πn

Solve

2 x - \frac{π}{6} = \frac{2 π}{3} + 2 πn : x = πn + \frac{5 π}{12}

Solve

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

x = πn + \frac{5 π}{12}, x = πn + \frac{3 π}{4}

Solutions for the range

\frac{π}{2} \leq x < π

x = \frac{3 π}{4}

\frac{( sin ( x ) )}{( 4.2 )} = \frac{( sin ( 9 5 ^{\circ} ) )}{( 10.26 )}

- sin (x) = - cos (x)

t, f = sin (k t)

6 sin (θ) = 3 + 3 sin (θ)

sin (θ) = \frac{- 2}{2 2}, 0 \leq θ < 2 π

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