What is the general solution for cot^2(θ)+csc(θ)=1,0<= θ<2pi ?

Q: What is the general solution for cot^2(θ)+csc(θ)=1,0<= θ<2pi ?

The general solution for cot^2(θ)+csc(θ)=1,0<= θ<2pi is θ= pi/2 ,θ=(7pi)/6 ,θ=(11pi)/6

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cot^2(θ)+csc(θ)=1,0<= θ<2pi

Solution

cot^{2} (θ) + csc (θ) = 1, 0 \leq θ < 2 π

Solution

θ = \frac{π}{2}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Degrees

θ = 9 0^{\circ}, θ = 21 0^{\circ}, θ = 33 0^{\circ}

Solution steps

cot^{2} (θ) + csc (θ) = 1, 0 \leq θ < 2 π

Subtract

1

from both sides

cot^{2} (θ) + csc (θ) - 1 = 0

Rewrite using trig identities

- 1 + cot^{2} (θ) + csc (θ)

Use the Pythagorean identity:

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

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

= - 1 + csc^{2} (θ) - 1 + csc (θ)

Simplify

- 1 + csc^{2} (θ) - 1 + csc (θ) : csc^{2} (θ) + csc (θ) - 2

- 1 + csc^{2} (θ) - 1 + csc (θ)

Group like terms

= csc^{2} (θ) + csc (θ) - 1 - 1

Subtract the numbers:

- 1 - 1 = - 2

= csc^{2} (θ) + csc (θ) - 2

= csc^{2} (θ) + csc (θ) - 2

- 2 + csc (θ) + csc^{2} (θ) = 0

Solve by substitution

- 2 + csc (θ) + csc^{2} (θ) = 0

Let:

csc (θ) = u

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

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

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

Write in the standard form

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

u^{2} + u - 2 = 0

Solve with the quadratic formula

u^{2} + u - 2 = 0

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 \cdot 1 \cdot (- 2)

Apply rule

- (- a) = a

= 1 + 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 \cdot 1}, u_{2} = \frac{- 1 - 3}{2 \cdot 1}

u = \frac{- 1 + 3}{2 \cdot 1} : 1

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

Add/Subtract the numbers:

- 1 + 3 = 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 - 3}{2 \cdot 1} : - 2

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

Subtract the numbers:

- 1 - 3 = - 4

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

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 4}{2}

Apply the fraction rule:

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

= - \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= - 2

The solutions to the quadratic equation are:

u = 1, u = - 2

Substitute back

u = csc (θ)

csc (θ) = 1, csc (θ) = - 2

csc (θ) = 1, csc (θ) = - 2

csc (θ) = 1, 0 \leq θ < 2 π : θ = \frac{π}{2}

csc (θ) = 1, 0 \leq θ < 2 π

General solutions for

csc (θ) = 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

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{2}

csc (θ) = - 2, 0 \leq θ < 2 π : θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

csc (θ) = - 2, 0 \leq θ < 2 π

General solutions for

csc (θ) = - 2

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

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

θ = \frac{7 π}{6} + 2 πn, θ = \frac{11 π}{6} + 2 πn

Solutions for the range

0 \leq θ < 2 π

θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

Combine all the solutions

θ = \frac{π}{2}, θ = \frac{7 π}{6}, θ = \frac{11 π}{6}

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tan^2(45-x/3)=0.58

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Frequently Asked Questions (FAQ)

What is the general solution for cot^2(θ)+csc(θ)=1,0<= θ<2pi ?
The general solution for cot^2(θ)+csc(θ)=1,0<= θ<2pi is θ= pi/2 ,θ=(7pi)/6 ,θ=(11pi)/6