What is the general solution for csc^2(θ)=3cot(θ)+5 ?

The general solution for csc^2(θ)=3cot(θ)+5 is θ=0.24497…+pin,θ=(3pi)/4+pin

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csc^2(θ)=3cot(θ)+5

Solution

csc^{2} (θ) = 3 cot (θ) + 5

Solution

θ = 0.24497 \dots + πn, θ = \frac{3 π}{4} + πn

Degrees

θ = 14.03624 \dots^{\circ} + 18 0^{\circ} n, θ = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

csc^{2} (θ) = 3 cot (θ) + 5

Subtract

3 cot (θ) + 5

from both sides

csc^{2} (θ) - 3 cot (θ) - 5 = 0

Rewrite using trig identities

- 5 + csc^{2} (θ) - 3 cot (θ)

Use the Pythagorean identity:

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

= - 5 + 1 + cot^{2} (θ) - 3 cot (θ)

Simplify

= cot^{2} (θ) - 3 cot (θ) - 4

- 4 + cot^{2} (θ) - 3 cot (θ) = 0

Solve by substitution

- 4 + cot^{2} (θ) - 3 cot (θ) = 0

Let:

cot (θ) = u

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

- 4 + u^{2} - 3 u = 0 : u = 4, u = - 1

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

Write in the standard form

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

u^{2} - 3 u - 4 = 0

Solve with the quadratic formula

u^{2} - 3 u - 4 = 0

Quadratic Equation Formula:

For

a = 1, b = - 3, c = - 4

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

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

(- 3)^{2} - 4 \cdot 1 \cdot (- 4) = 5

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

Apply rule

- (- a) = a

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

Apply exponent rule:

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

n

is even

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

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

Multiply the numbers:

4 \cdot 1 \cdot 4 = 16

= 3^{2} + 16

3^{2} = 9

= 9 + 16

Add the numbers:

9 + 16 = 25

= 25

Factor the number:

25 = 5^{2}

= 5^{2}

Apply radical rule:

5^{2} = 5

= 5

u_{1, 2} = \frac{- ( - 3 ) \pm 5}{2 \cdot 1}

Separate the solutions

u_{1} = \frac{- ( - 3 ) + 5}{2 \cdot 1}, u_{2} = \frac{- ( - 3 ) - 5}{2 \cdot 1}

u = \frac{- ( - 3 ) + 5}{2 \cdot 1} : 4

\frac{- ( - 3 ) + 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 + 5}{2 \cdot 1}

Add the numbers:

3 + 5 = 8

= \frac{8}{2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{8}{2}

Divide the numbers:

\frac{8}{2} = 4

= 4

u = \frac{- ( - 3 ) - 5}{2 \cdot 1} : - 1

\frac{- ( - 3 ) - 5}{2 \cdot 1}

Apply rule

- (- a) = a

= \frac{3 - 5}{2 \cdot 1}

Subtract the numbers:

3 - 5 = - 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 = 4, u = - 1

Substitute back

u = cot (θ)

cot (θ) = 4, cot (θ) = - 1

cot (θ) = 4, cot (θ) = - 1

cot (θ) = 4 : θ = arccot (4) + πn

cot (θ) = 4

Apply trig inverse properties

cot (θ) = 4

General solutions for

cot (θ) = 4

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

θ = arccot (4) + πn

θ = arccot (4) + πn

cot (θ) = - 1 : θ = \frac{3 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{3 π}{4} + πn

θ = \frac{3 π}{4} + πn

Combine all the solutions

θ = arccot (4) + πn, θ = \frac{3 π}{4} + πn

Show solutions in decimal form

θ = 0.24497 \dots + πn, θ = \frac{3 π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for csc^2(θ)=3cot(θ)+5 ?
The general solution for csc^2(θ)=3cot(θ)+5 is θ=0.24497…+pin,θ=(3pi)/4+pin