What is the general solution for cos(2x-pi/6)=((sqrt(3)))/2 ,0<x<2pi ?

The general solution for cos(2x-pi/6)=((sqrt(3)))/2 ,0<x<2pi is x= pi/6 ,x=pi,x=(7pi)/6

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cos(2x-pi/6)=((sqrt(3)))/2 ,0<x<2pi

cos (2 x - \frac{π}{6}) = \frac{( 3 )}{2}, 0 < x < 2 π

x = \frac{π}{6}, x = π, x = \frac{7 π}{6}

Solution steps

cos (2 x - \frac{π}{6}) = \frac{( 3 )}{2}, 0 < x < 2 π

General solutions for

cos (2 x - \frac{π}{6}) = \frac{3}{2}

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

Solve

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

Solve

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

x = \frac{π}{6} + πn, x = πn + π

Solutions for the range

0 < x < 2 π

x = \frac{π}{6}, x = π, x = \frac{7 π}{6}

- 1 = tan (x)

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

2 tan (x) = 1 - tan^{2} (x)

3 sin (θ) + 3 = 2 cos^{2} (θ)

cos (x) = - \frac{5}{6}

What is the general solution for cos(2x-pi/6)=((sqrt(3)))/2 ,0<x<2pi ?
The general solution for cos(2x-pi/6)=((sqrt(3)))/2 ,0<x<2pi is x= pi/6 ,x=pi,x=(7pi)/6