What is the general solution for tan(θ/2-pi/6)=-1,0<= θ<2pi ?

The general solution for tan(θ/2-pi/6)=-1,0<= θ<2pi is θ=(11pi)/6

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tan(θ/2-pi/6)=-1,0<= θ<2pi

tan (\frac{θ}{2} - \frac{π}{6}) = - 1, 0 \leq θ < 2 π

θ = \frac{11 π}{6}

Solution steps

tan (\frac{θ}{2} - \frac{π}{6}) = - 1, 0 \leq θ < 2 π

General solutions for

tan (\frac{θ}{2} - \frac{π}{6}) = - 1

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

Solve

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

θ = 2 πn + \frac{11 π}{6}

Solutions for the range

0 \leq θ < 2 π

θ = \frac{11 π}{6}

arctan (x + 2) - arctan (x + 1) = \frac{π}{4}

3 cos (2 x) = 1

cos^{2} (x) = (1 + sin (x)) (1 - cos (x))

tan (x) = 1, - π < x \leq π

\frac{1}{2 cos ^{2} ( x - 1 )} = \frac{1 + tan ^{2} ( x )}{2 sec ^{2} ( x )}

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