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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = \frac{π}{2} + πn, x = \frac{3 π}{4} + πn

Degrees

x = 9 0^{\circ} + 18 0^{\circ} n, x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

1

from both sides

csc^{2} (x) + cot (x) - 1 = 0

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

cot (x) + cot^{2} (x) = 0

Solve by substitution

cot (x) + cot^{2} (x) = 0

Let:

cot (x) = u

u + u^{2} = 0

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

u + u^{2} = 0

Write in the standard form

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

u^{2} + u = 0

Solve with the quadratic formula

u^{2} + u = 0

Quadratic Equation Formula:

For

a = 1, b = 1, c = 0

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

1^{2} - 4 \cdot 1 \cdot 0 = 1

1^{2} - 4 \cdot 1 \cdot 0

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- 1 \pm 1}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 1}{2 \cdot 1}, u_{2} = \frac{- 1 - 1}{2 \cdot 1}

u = \frac{- 1 + 1}{2 \cdot 1} : 0

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

Add/Subtract the numbers:

- 1 + 1 = 0

= \frac{0}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{0}{2}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

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

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

Subtract the numbers:

- 1 - 1 = - 2

= \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

The solutions to the quadratic equation are:

u = 0, u = - 1

Substitute back

u = cot (x)

cot (x) = 0, cot (x) = - 1

cot (x) = 0, cot (x) = - 1

cot (x) = 0 : x = \frac{π}{2} + πn

cot (x) = 0

General solutions for

cot (x) = 0

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}{2} + πn

x = \frac{π}{2} + πn

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{π}{2} + πn, x = \frac{3 π}{4} + πn

Popular Examples

cos(θ)=-1/6

cos (θ) = - \frac{1}{6}

arcsin(x)=1

arcsin (x) = 1

sin(θ)=(sqrt(5))/3

sin (θ) = \frac{5}{3}

cos((3x)/2)=0

cos (\frac{3 x}{2}) = 0

sec(x)+1=tan(x)

sec (x) + 1 = tan (x)

Frequently Asked Questions (FAQ)

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