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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

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

Subtract

csc (x)

from both sides

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= - csc (x) + csc^{2} (x)

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

Solve by substitution

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

Let:

csc (x) = u

- u + u^{2} = 0

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

- 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

(- 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 1 \cdot 0 = 0

4 \cdot 1 \cdot 0

Apply rule

0 \cdot a = 0

= 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} : 1

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

Apply rule

- (- a) = a

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

Add the numbers:

1 + 1 = 2

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

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

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

Apply rule

- (- a) = a

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

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

The solutions to the quadratic equation are:

u = 1, u = 0

Substitute back

u = csc (x)

csc (x) = 1, csc (x) = 0

csc (x) = 1, csc (x) = 0

csc (x) = 1 : x = \frac{π}{2} + 2 πn

csc (x) = 1

General solutions for

csc (x) = 1

csc (x)

periodicity table with

2 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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

csc (x) = 0 :

No Solution

csc (x) = 0

csc (x) \leq - 1 or csc (x) \geq 1

No Solution

Combine all the solutions

x = \frac{π}{2} + 2 πn

Popular Examples

cos(θ)= 7/(3sqrt(17))

cos (θ) = \frac{7}{3 17}

sin(A)-0.1*cos(A)=(6.94)/(9.8)

sin (A) - 0.1 \cdot cos (A) = \frac{6.94}{9.8}

solvefor t,s=2arctan(t)solve for

t, s = 2 arctan (t)

(cos(x))/(tan(x))= 3/2

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

cos(2x)+5cos(x)=3

cos (2 x) + 5 cos (x) = 3

Frequently Asked Questions (FAQ)

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