What is the general solution for 3cot(θ)-1=-cot^2(θ) ?

The general solution for 3cot(θ)-1=-cot^2(θ) is θ=2.84759…+pin,θ=1.27679…+pin

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Popular Trigonometry >

3cot(θ)-1=-cot^2(θ)

Solution

3 cot (θ) - 1 = - cot^{2} (θ)

Solution

θ = 2.84759 \dots + πn, θ = 1.27679 \dots + πn

Degrees

θ = 163.15496 \dots^{\circ} + 18 0^{\circ} n, θ = 73.15496 \dots^{\circ} + 18 0^{\circ} n

Solution steps

3 cot (θ) - 1 = - cot^{2} (θ)

Solve by substitution

3 cot (θ) - 1 = - cot^{2} (θ)

Let:

cot (θ) = u

3 u - 1 = - u^{2}

3 u - 1 = - u^{2} : u = - \frac{3 + 13}{2}, u = \frac{13 - 3}{2}

3 u - 1 = - u^{2}

Switch sides

- u^{2} = 3 u - 1

Move

1

to the left side

- u^{2} = 3 u - 1

Add

1

to both sides

- u^{2} + 1 = 3 u - 1 + 1

Simplify

- u^{2} + 1 = 3 u

- u^{2} + 1 = 3 u

Move

3 u

to the left side

- u^{2} + 1 = 3 u

Subtract

3 u

from both sides

- u^{2} + 1 - 3 u = 3 u - 3 u

Simplify

- u^{2} + 1 - 3 u = 0

- u^{2} + 1 - 3 u = 0

Write in the standard form

a x^{2} + b x + c = 0

- u^{2} - 3 u + 1 = 0

Solve with the quadratic formula

- u^{2} - 3 u + 1 = 0

Quadratic Equation Formula:

For

a = - 1, b = - 3, c = 1

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

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

(- 3)^{2} - 4 (- 1) \cdot 1 = 13

(- 3)^{2} - 4 (- 1) \cdot 1

Apply rule

- (- a) = a

= (- 3)^{2} + 4 \cdot 1 \cdot 1

Apply exponent rule:

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

n

is even

(- 3)^{2} = 3^{2}

= 3^{2} + 4 \cdot 1 \cdot 1

Multiply the numbers:

4 \cdot 1 \cdot 1 = 4

= 3^{2} + 4

3^{2} = 9

= 9 + 4

Add the numbers:

9 + 4 = 13

= 13

u_{1, 2} = \frac{- ( - 3 ) \pm 13}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 13}{2 ( - 1 )}, u_{2} = \frac{- ( - 3 ) - 13}{2 ( - 1 )}

u = \frac{- ( - 3 ) + 13}{2 ( - 1 )} : - \frac{3 + 13}{2}

\frac{- ( - 3 ) + 13}{2 ( - 1 )}

Remove parentheses:

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

= \frac{3 + 13}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{3 + 13}{- 2}

Apply the fraction rule:

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

= - \frac{3 + 13}{2}

u = \frac{- ( - 3 ) - 13}{2 ( - 1 )} : \frac{13 - 3}{2}

\frac{- ( - 3 ) - 13}{2 ( - 1 )}

Remove parentheses:

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

= \frac{3 - 13}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{3 - 13}{- 2}

Apply the fraction rule:

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

3 - 13 = - (13 - 3)

= \frac{13 - 3}{2}

The solutions to the quadratic equation are:

u = - \frac{3 + 13}{2}, u = \frac{13 - 3}{2}

Substitute back

u = cot (θ)

cot (θ) = - \frac{3 + 13}{2}, cot (θ) = \frac{13 - 3}{2}

cot (θ) = - \frac{3 + 13}{2}, cot (θ) = \frac{13 - 3}{2}

cot (θ) = - \frac{3 + 13}{2} : θ = arccot (- \frac{3 + 13}{2}) + πn

cot (θ) = - \frac{3 + 13}{2}

Apply trig inverse properties

cot (θ) = - \frac{3 + 13}{2}

General solutions for

cot (θ) = - \frac{3 + 13}{2}

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

θ = arccot (- \frac{3 + 13}{2}) + πn

θ = arccot (- \frac{3 + 13}{2}) + πn

cot (θ) = \frac{13 - 3}{2} : θ = arccot (\frac{13 - 3}{2}) + πn

cot (θ) = \frac{13 - 3}{2}

Apply trig inverse properties

cot (θ) = \frac{13 - 3}{2}

General solutions for

cot (θ) = \frac{13 - 3}{2}

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

θ = arccot (\frac{13 - 3}{2}) + πn

θ = arccot (\frac{13 - 3}{2}) + πn

Combine all the solutions

θ = arccot (- \frac{3 + 13}{2}) + πn, θ = arccot (\frac{13 - 3}{2}) + πn

Show solutions in decimal form

θ = 2.84759 \dots + πn, θ = 1.27679 \dots + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 3cot(θ)-1=-cot^2(θ) ?
The general solution for 3cot(θ)-1=-cot^2(θ) is θ=2.84759…+pin,θ=1.27679…+pin