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

The general solution for -cot^2(x)=1+2cot(x) is x=(3pi)/4+pin

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

Solution

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

Solution

x = \frac{3 π}{4} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cot (x) = u

- u^{2} = 1 + 2 u

- u^{2} = 1 + 2 u : u = - 1

- u^{2} = 1 + 2 u

Move

2 u

to the left side

- u^{2} = 1 + 2 u

Subtract

2 u

from both sides

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

Simplify

- u^{2} - 2 u = 1

- u^{2} - 2 u = 1

Move

1

to the left side

- u^{2} - 2 u = 1

Subtract

1

from both sides

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

Simplify

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 1, b = - 2, c = - 1

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

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

(- 2)^{2} - 4 (- 1) (- 1) = 0

(- 2)^{2} - 4 (- 1) (- 1)

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

= 2^{2} - 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 2^{2} - 4

2^{2} = 4

= 4 - 4

Subtract the numbers:

4 - 4 = 0

= 0

u_{1, 2} = \frac{- ( - 2 ) \pm 0}{2 ( - 1 )}

u = \frac{- ( - 2 )}{2 ( - 1 )}

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

\frac{- ( - 2 )}{2 ( - 1 )}

Remove parentheses:

(- a) = - a, - (- a) = a

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{- 2}

Apply the fraction rule:

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

= - \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= - 1

u = - 1

The solution to the quadratic equation is:

u = - 1

Substitute back

u = cot (x)

cot (x) = - 1

cot (x) = - 1

cot (x) = - 1 : x = \frac{3 π}{4} + πn

cot (x) = - 1

General solutions for

cot (x) = - 1

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{3 π}{4} + πn

x = \frac{3 π}{4} + πn

Combine all the solutions

x = \frac{3 π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for -cot^2(x)=1+2cot(x) ?
The general solution for -cot^2(x)=1+2cot(x) is x=(3pi)/4+pin