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}

Degrees

t = 22. 5^{\circ}, t = 112. 5^{\circ}, t = 202. 5^{\circ}, t = 292. 5^{\circ}

Solution steps

tan (2 t) = 1, 0 \leq t < 2 π

General solutions for

tan (2 t) = 1

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

2 t = \frac{π}{4} + πn

2 t = \frac{π}{4} + πn

Solve

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

2 t = \frac{π}{4} + πn

Divide both sides by

2

2 t = \frac{π}{4} + πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 t}{2} : t

\frac{2 t}{2}

Divide the numbers:

\frac{2}{2} = 1

= t

Simplify

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

\frac{\frac{π}{4}}{2} + \frac{πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

= \frac{π}{8} + \frac{πn}{2}

t = \frac{π}{8} + \frac{πn}{2}

t = \frac{π}{8} + \frac{πn}{2}

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