What is the general solution for tan(2t)=1 ?

The general solution for tan(2t)=1 is t= pi/8+(pin)/2

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tan(2t)=1

tan (2 t) = 1

t = \frac{π}{8} + \frac{πn}{2}

Degrees

t = 22. 5^{\circ} + 9 0^{\circ} n

Solution steps

tan (2 t) = 1

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}

21 = 3 \cdot 2 \cdot 25 sin (x) + 9.8 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 0.5

sin (2 x) = \frac{( 3 m - 8 )}{6}

cos (x) (18 - 122) = 12 + 63

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

cos (3 x + 20) = - sin (x - 6 0^{\circ})

What is the general solution for tan(2t)=1 ?
The general solution for tan(2t)=1 is t= pi/8+(pin)/2