What is the general solution for tan(2t)=1,0<= t<2pi ?

The general solution for tan(2t)=1,0<= t<2pi is t= pi/8 ,t=(5pi)/8 ,t=(9pi)/8 ,t=(13pi)/8

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tan(2t)=1,0<= t<2pi

tan (2 t) = 1, 0 \leq t < 2 π

t = \frac{π}{8}, t = \frac{5 π}{8}, t = \frac{9 π}{8}, t = \frac{13 π}{8}

Solution steps

tan (2 t) = 1, 0 \leq t < 2 π

General solutions for

tan (2 t) = 1

2 t = \frac{π}{4} + πn

Solve

2 t = \frac{π}{4} + πn : t = \frac{π}{8} + \frac{πn}{2}

t = \frac{π}{8} + \frac{πn}{2}

Solutions for the range

0 \leq t < 2 π

t = \frac{π}{8}, t = \frac{5 π}{8}, t = \frac{9 π}{8}, t = \frac{13 π}{8}

sin^{2} (x) = cot (x)

sin (\frac{x}{2}) = 1, 0 \leq x < 2 π

sin (\frac{x}{2}) = - cos (x - \frac{π}{3})

3 = cot^{2} (θ) + 4 cot (θ)

cos^{2} (2 x) + 3 sin (2 x) - 3 = 0

What is the general solution for tan(2t)=1,0<= t<2pi ?
The general solution for tan(2t)=1,0<= t<2pi is t= pi/8 ,t=(5pi)/8 ,t=(9pi)/8 ,t=(13pi)/8