What is the general solution for sqrt(3)cot^2(x)-4cot(x)+sqrt(3)=0 ?

The general solution for sqrt(3)cot^2(x)-4cot(x)+sqrt(3)=0 is x= pi/6+pin,x= pi/3+pin

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sqrt(3)cot^2(x)-4cot(x)+sqrt(3)=0

Solution

3 cot^{2} (x) - 4 cot (x) + 3 = 0

Solution

x = \frac{π}{6} + πn, x = \frac{π}{3} + πn

Degrees

x = 3 0^{\circ} + 18 0^{\circ} n, x = 6 0^{\circ} + 18 0^{\circ} n

Solution steps

3 cot^{2} (x) - 4 cot (x) + 3 = 0

Solve by substitution

3 cot^{2} (x) - 4 cot (x) + 3 = 0

Let:

cot (x) = u

3 u^{2} - 4 u + 3 = 0

3 u^{2} - 4 u + 3 = 0 : u = 3, u = \frac{3}{3}

3 u^{2} - 4 u + 3 = 0

Solve with the quadratic formula

3 u^{2} - 4 u + 3 = 0

Quadratic Equation Formula:

For

a = 3, b = - 4, c = 3

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 3 3}{2 3}

u_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 3 3}{2 3}

(- 4)^{2} - 433 = 2

(- 4)^{2} - 433

(- 4)^{2} = 4^{2}

(- 4)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 4)^{2} = 4^{2}

= 4^{2}

433 = 12

433

Apply radical rule:

a a = a

33 = 3

= 4 \cdot 3

Multiply the numbers:

4 \cdot 3 = 12

= 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

Subtract the numbers:

16 - 12 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u_{1, 2} = \frac{- ( - 4 ) \pm 2}{2 3}

Separate the solutions

u_{1} = \frac{- ( - 4 ) + 2}{2 3}, u_{2} = \frac{- ( - 4 ) - 2}{2 3}

u = \frac{- ( - 4 ) + 2}{2 3} : 3

\frac{- ( - 4 ) + 2}{2 3}

Apply rule

- (- a) = a

= \frac{4 + 2}{2 3}

Add the numbers:

4 + 2 = 6

= \frac{6}{2 3}

Divide the numbers:

\frac{6}{2} = 3

= \frac{3}{3}

Apply radical rule:

3 = 3^{\frac{1}{2}}

= \frac{3}{3 ^{\frac{1}{2}}}

Apply exponent rule:

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

\frac{3 ^{1}}{3 ^{\frac{1}{2}}} = 3^{1 - \frac{1}{2}}

= 3^{1 - \frac{1}{2}}

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

= 3^{\frac{1}{2}}

Apply radical rule:

3^{\frac{1}{2}} = 3

= 3

u = \frac{- ( - 4 ) - 2}{2 3} : \frac{3}{3}

\frac{- ( - 4 ) - 2}{2 3}

Apply rule

- (- a) = a

= \frac{4 - 2}{2 3}

Subtract the numbers:

4 - 2 = 2

= \frac{2}{2 3}

Divide the numbers:

\frac{2}{2} = 1

= \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}

The solutions to the quadratic equation are:

u = 3, u = \frac{3}{3}

Substitute back

u = cot (x)

cot (x) = 3, cot (x) = \frac{3}{3}

cot (x) = 3, cot (x) = \frac{3}{3}

cot (x) = 3 : x = \frac{π}{6} + πn

cot (x) = 3

General solutions for

cot (x) = 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

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

cot (x) = \frac{3}{3} : x = \frac{π}{3} + πn

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

General solutions for

cot (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

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

Combine all the solutions

x = \frac{π}{6} + πn, x = \frac{π}{3} + πn

Popular Examples

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3cos^2(θ)-1=0

3 cos^{2} (θ) - 1 = 0

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sin (3 x - \frac{π}{6}) = - cos (3 x - \frac{π}{6})

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cos^{2} (x) = \frac{1}{2} (1 + cos (x))

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)cot^2(x)-4cot(x)+sqrt(3)=0 ?
The general solution for sqrt(3)cot^2(x)-4cot(x)+sqrt(3)=0 is x= pi/6+pin,x= pi/3+pin