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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-2cot(θ)=-1-cot^2(θ)

Solution

- 2 cot (θ) = - 1 - cot^{2} (θ)

Solution

θ = \frac{π}{4} + πn

Degrees

θ = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

- 2 cot (θ) = - 1 - cot^{2} (θ)

Solve by substitution

- 2 cot (θ) = - 1 - cot^{2} (θ)

Let:

cot (θ) = u

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

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

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

Switch sides

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

Move

2 u

to the left side

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

Add

2 u

to both sides

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

Simplify

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

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

Write in the standard form

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

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

= \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 (θ)

cot (θ) = 1

cot (θ) = 1

cot (θ) = 1 : θ = \frac{π}{4} + πn

cot (θ) = 1

General solutions for

cot (θ) = 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

θ = \frac{π}{4} + πn

θ = \frac{π}{4} + πn

Combine all the solutions

θ = \frac{π}{4} + πn

Popular Examples

-cos^2(x)-sin(x)-1=0

- cos^{2} (x) - sin (x) - 1 = 0

-8cos^2(x)+1=4sin^2(x)-8cos(x)

- 8 cos^{2} (x) + 1 = 4 sin^{2} (x) - 8 cos (x)

(1+csc(y))/(cot(y)+cos(y))=csc(y)

\frac{1 + csc ( y )}{cot ( y ) + cos ( y )} = csc (y)

2=sin(10x)

2 = sin (10 x)

cos(2x)=((2k-8))/9

cos (2 x) = \frac{( 2 k - 8 )}{9}

Frequently Asked Questions (FAQ)

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