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} < x < π

x = \frac{3 π}{4}

Solution steps

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

x = \frac{3 π}{4}

sin (2 θ) + sin (θ) + 2 cos (θ) + 1 = 0

tan (θ) = \frac{563}{2000}

cot (\frac{x}{3}) = 1

tan (t) = 7

cos (x) + cos (x) = 0

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