What is the general solution for tan(2x)-1=0 ?

The general solution for tan(2x)-1=0 is x= pi/8+(pin)/2

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tan(2x)-1=0

tan (2 x) - 1 = 0

x = \frac{π}{8} + \frac{πn}{2}

Degrees

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

Solution steps

tan (2 x) - 1 = 0

Move

1

to the right side

tan (2 x) - 1 = 0

Add

1

to both sides

tan (2 x) - 1 + 1 = 0 + 1

Simplify

tan (2 x) = 1

tan (2 x) = 1

General solutions for

tan (2 x) = 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 x = \frac{π}{4} + πn

2 x = \frac{π}{4} + πn

Solve

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

2 x = \frac{π}{4} + πn

Divide both sides by

2

2 x = \frac{π}{4} + πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

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}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

tan (x) = - \frac{1}{3}

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

sin (x) - cos (2 x) = 0

csc (x) = - 2

cos (2 t) = sin (t)

What is the general solution for tan(2x)-1=0 ?
The general solution for tan(2x)-1=0 is x= pi/8+(pin)/2