What is the general solution for 12cot^2(x)-cot(x)=1 ?

The general solution for 12cot^2(x)-cot(x)=1 is x=1.24904…+pin,x=1.81577…+pin

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12cot^2(x)-cot(x)=1

Solution

12 cot^{2} (x) - cot (x) = 1

Solution

x = 1.24904 \dots + πn, x = 1.81577 \dots + πn

Degrees

x = 71.56505 \dots^{\circ} + 18 0^{\circ} n, x = 104.03624 \dots^{\circ} + 18 0^{\circ} n

Solution steps

12 cot^{2} (x) - cot (x) = 1

Solve by substitution

12 cot^{2} (x) - cot (x) = 1

Let:

cot (x) = u

12 u^{2} - u = 1

12 u^{2} - u = 1 : u = \frac{1}{3}, u = - \frac{1}{4}

12 u^{2} - u = 1

Move

1

to the left side

12 u^{2} - u = 1

Subtract

1

from both sides

12 u^{2} - u - 1 = 1 - 1

Simplify

12 u^{2} - u - 1 = 0

12 u^{2} - u - 1 = 0

Solve with the quadratic formula

12 u^{2} - u - 1 = 0

Quadratic Equation Formula:

For

a = 12, b = - 1, c = - 1

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 12 ( - 1 )}{2 \cdot 12}

u_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 12 ( - 1 )}{2 \cdot 12}

(- 1)^{2} - 4 \cdot 12 (- 1) = 7

(- 1)^{2} - 4 \cdot 12 (- 1)

Apply rule

- (- a) = a

= (- 1)^{2} + 4 \cdot 12 \cdot 1

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 12 \cdot 1 = 48

4 \cdot 12 \cdot 1

Multiply the numbers:

4 \cdot 12 \cdot 1 = 48

= 48

= 1 + 48

Add the numbers:

1 + 48 = 49

= 49

Factor the number:

49 = 7^{2}

= 7^{2}

Apply radical rule:

7^{2} = 7

= 7

u_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 \cdot 12}

Separate the solutions

u_{1} = \frac{- ( - 1 ) + 7}{2 \cdot 12}, u_{2} = \frac{- ( - 1 ) - 7}{2 \cdot 12}

u = \frac{- ( - 1 ) + 7}{2 \cdot 12} : \frac{1}{3}

\frac{- ( - 1 ) + 7}{2 \cdot 12}

Apply rule

- (- a) = a

= \frac{1 + 7}{2 \cdot 12}

Add the numbers:

1 + 7 = 8

= \frac{8}{2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{8}{24}

Cancel the common factor:

8

= \frac{1}{3}

u = \frac{- ( - 1 ) - 7}{2 \cdot 12} : - \frac{1}{4}

\frac{- ( - 1 ) - 7}{2 \cdot 12}

Apply rule

- (- a) = a

= \frac{1 - 7}{2 \cdot 12}

Subtract the numbers:

1 - 7 = - 6

= \frac{- 6}{2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{- 6}{24}

Apply the fraction rule:

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

= - \frac{6}{24}

Cancel the common factor:

6

= - \frac{1}{4}

The solutions to the quadratic equation are:

u = \frac{1}{3}, u = - \frac{1}{4}

Substitute back

u = cot (x)

cot (x) = \frac{1}{3}, cot (x) = - \frac{1}{4}

cot (x) = \frac{1}{3}, cot (x) = - \frac{1}{4}

cot (x) = \frac{1}{3} : x = arccot (\frac{1}{3}) + πn

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

Apply trig inverse properties

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

General solutions for

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

cot (x) = a \Rightarrow x = arccot (a) + πn

x = arccot (\frac{1}{3}) + πn

x = arccot (\frac{1}{3}) + πn

cot (x) = - \frac{1}{4} : x = arccot (- \frac{1}{4}) + πn

cot (x) = - \frac{1}{4}

Apply trig inverse properties

cot (x) = - \frac{1}{4}

General solutions for

cot (x) = - \frac{1}{4}

cot (x) = - a \Rightarrow x = arccot (- a) + πn

x = arccot (- \frac{1}{4}) + πn

x = arccot (- \frac{1}{4}) + πn

Combine all the solutions

x = arccot (\frac{1}{3}) + πn, x = arccot (- \frac{1}{4}) + πn

Show solutions in decimal form

x = 1.24904 \dots + πn, x = 1.81577 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 12cot^2(x)-cot(x)=1 ?
The general solution for 12cot^2(x)-cot(x)=1 is x=1.24904…+pin,x=1.81577…+pin