What is the general solution for cot(2x)=0 ?

The general solution for cot(2x)=0 is x= pi/4+(pin)/2

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cot(2x)=0

cot (2 x) = 0

x = \frac{π}{4} + \frac{πn}{2}

Degrees

x = 4 5^{\circ} + 9 0^{\circ} n

Solution steps

cot (2 x) = 0

General solutions for

cot (2 x) = 0

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

2 x = \frac{π}{2} + πn

2 x = \frac{π}{2} + πn

Solve

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

2 x = \frac{π}{2} + πn

Divide both sides by

2

2 x = \frac{π}{2} + πn

Divide both sides by

2

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

Simplify

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

\frac{\frac{π}{2}}{2} + \frac{πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

= \frac{π}{4} + \frac{πn}{2}

x = \frac{π}{4} + \frac{πn}{2}

x = \frac{π}{4} + \frac{πn}{2}

x = \frac{π}{4} + \frac{πn}{2}

x = \frac{π}{4} + \frac{πn}{2}

cos (2 x) = cos^{2} (x)

tan (\frac{x}{2}) = \frac{3}{3}

sec (x) csc (x) + 2 csc (x) = 0

sin (x) + 3 cos (x) = 1

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

What is the general solution for cot(2x)=0 ?
The general solution for cot(2x)=0 is x= pi/4+(pin)/2