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

Divide both sides by

- 1

- cot (2 x) = 0

Divide both sides by

- 1

\frac{- cot ( 2 x )}{- 1} = \frac{0}{- 1}

Simplify

cot (2 x) = 0

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 (x) = \frac{π}{6}

3 cos (2 θ) + 6 = - 5 cos (θ) + 2

sin (3 x) cos (x) - sin (x) cos (3 x) = 0

sin (x) = \frac{8}{12}

2 cos (θ) sin (θ) + 3 cos (θ) = 0

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