What is the general solution for sqrt(3)cot(2x)=-1,0<= x<= pi ?

Q: What is the general solution for sqrt(3)cot(2x)=-1,0<= x<= pi ?

The general solution for sqrt(3)cot(2x)=-1,0<= x<= pi is x= pi/3 ,x=(5pi)/6

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sqrt(3)cot(2x)=-1,0<= x<= pi

Solution

3 cot (2 x) = - 1, 0 \leq x \leq π

Solution

x = \frac{π}{3}, x = \frac{5 π}{6}

Degrees

x = 6 0^{\circ}, x = 15 0^{\circ}

Solution steps

3 cot (2 x) = - 1, 0 \leq x \leq π

Divide both sides by

3

3 cot (2 x) = - 1

Divide both sides by

3

\frac{3 cot ( 2 x )}{3} = \frac{- 1}{3}

Simplify

\frac{3 cot ( 2 x )}{3} = \frac{- 1}{3}

Simplify

\frac{3 cot ( 2 x )}{3} : cot (2 x)

\frac{3 cot ( 2 x )}{3}

Cancel the common factor:

3

= cot (2 x)

Simplify

\frac{- 1}{3} : - \frac{3}{3}

\frac{- 1}{3}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{3}

Rationalize

- \frac{1}{3} : - \frac{3}{3}

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= - \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

cot (2 x) = - \frac{3}{3}

cot (2 x) = - \frac{3}{3}

cot (2 x) = - \frac{3}{3}

General solutions for

cot (2 x) = - \frac{3}{3}

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{2 π}{3} + πn

2 x = \frac{2 π}{3} + πn

Solve

2 x = \frac{2 π}{3} + πn : x = \frac{π}{3} + \frac{πn}{2}

2 x = \frac{2 π}{3} + πn

Divide both sides by

2

2 x = \frac{2 π}{3} + πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{2 π}{3}}{2} + \frac{πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{2 π}{3}}{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{2 π}{3}}{2} + \frac{πn}{2} : \frac{π}{3} + \frac{πn}{2}

\frac{\frac{2 π}{3}}{2} + \frac{πn}{2}

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{2 π}{6}

Cancel the common factor:

2

= \frac{π}{3}

= \frac{π}{3} + \frac{πn}{2}

x = \frac{π}{3} + \frac{πn}{2}

x = \frac{π}{3} + \frac{πn}{2}

x = \frac{π}{3} + \frac{πn}{2}

x = \frac{π}{3} + \frac{πn}{2}

Solutions for the range

0 \leq x \leq π

x = \frac{π}{3}, x = \frac{5 π}{6}

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)cot(2x)=-1,0<= x<= pi ?
The general solution for sqrt(3)cot(2x)=-1,0<= x<= pi is x= pi/3 ,x=(5pi)/6