What is the general solution for -csc(θ)+5=cot(θ)+6 ?

The general solution for -csc(θ)+5=cot(θ)+6 is θ=(3pi)/2+2pin

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-csc(θ)+5=cot(θ)+6

Solution

- csc (θ) + 5 = cot (θ) + 6

Solution

θ = \frac{3 π}{2} + 2 πn

Degrees

θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

- csc (θ) + 5 = cot (θ) + 6

Subtract

cot (θ) + 6

from both sides

- csc (θ) - cot (θ) - 1 = 0

Express with sin, cos

- \frac{1}{sin ( θ )} - \frac{cos ( θ )}{sin ( θ )} - 1 = 0

Simplify

- \frac{1}{sin ( θ )} - \frac{cos ( θ )}{sin ( θ )} - 1 : \frac{- 1 - cos ( θ ) - sin ( θ )}{sin ( θ )}

- \frac{1}{sin ( θ )} - \frac{cos ( θ )}{sin ( θ )} - 1

Combine the fractions

- \frac{1}{sin ( θ )} - \frac{cos ( θ )}{sin ( θ )} : \frac{- 1 - cos ( θ )}{sin ( θ )}

Apply rule

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 - cos ( θ )}{sin ( θ )}

= \frac{- cos ( θ ) - 1}{sin ( θ )} - 1

Convert element to fraction:

1 = \frac{1 sin ( θ )}{sin ( θ )}

= \frac{- 1 - cos ( θ )}{sin ( θ )} - \frac{1 \cdot sin ( θ )}{sin ( θ )}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{- 1 - cos ( θ ) - 1 \cdot sin ( θ )}{sin ( θ )}

Multiply:

1 \cdot sin (θ) = sin (θ)

= \frac{- 1 - cos ( θ ) - sin ( θ )}{sin ( θ )}

\frac{- 1 - cos ( θ ) - sin ( θ )}{sin ( θ )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 1 - cos (θ) - sin (θ) = 0

Add

sin (θ)

to both sides

- 1 - cos (θ) = sin (θ)

Square both sides

(- 1 - cos (θ))^{2} = sin^{2} (θ)

Subtract

sin^{2} (θ)

from both sides

(- 1 - cos (θ))^{2} - sin^{2} (θ) = 0

Rewrite using trig identities

(- 1 - cos (θ))^{2} - sin^{2} (θ)

Use the Pythagorean identity:

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

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

= (- 1 - cos (θ))^{2} - (1 - cos^{2} (θ))

Simplify

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

(- 1 - cos (θ))^{2} - (1 - cos^{2} (θ))

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

Apply Perfect Square Formula:

(a - b)^{2} = a^{2} - 2 ab + b^{2}

a = - 1, b = cos (θ)

= (- 1)^{2} - 2 (- 1) cos (θ) + cos^{2} (θ)

Simplify

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

(- 1)^{2} - 2 (- 1) cos (θ) + cos^{2} (θ)

Apply rule

- (- a) = a

= (- 1)^{2} + 2 \cdot 1 \cdot cos (θ) + cos^{2} (θ)

(- 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

2 \cdot 1 \cdot cos (θ) = 2 cos (θ)

2 \cdot 1 \cdot cos (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos (θ)

= 1 + 2 cos (θ) + cos^{2} (θ)

= 1 + 2 cos (θ) + cos^{2} (θ)

= 1 + 2 cos (θ) + cos^{2} (θ) - (1 - cos^{2} (θ))

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

- (1 - cos^{2} (θ))

Distribute parentheses

= - (1) - (- cos^{2} (θ))

Apply minus-plus rules

- (- a) = a, - (a) = - a

= - 1 + cos^{2} (θ)

= 1 + 2 cos (θ) + cos^{2} (θ) - 1 + cos^{2} (θ)

Simplify

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

1 + 2 cos (θ) + cos^{2} (θ) - 1 + cos^{2} (θ)

Group like terms

= 2 cos (θ) + cos^{2} (θ) + cos^{2} (θ) + 1 - 1

Add similar elements:

cos^{2} (θ) + cos^{2} (θ) = 2 cos^{2} (θ)

= 2 cos (θ) + 2 cos^{2} (θ) + 1 - 1

1 - 1 = 0

= 2 cos^{2} (θ) + 2 cos (θ)

= 2 cos^{2} (θ) + 2 cos (θ)

= 2 cos^{2} (θ) + 2 cos (θ)

2 cos (θ) + 2 cos^{2} (θ) = 0

Solve by substitution

2 cos (θ) + 2 cos^{2} (θ) = 0

Let:

cos (θ) = u

2 u + 2 u^{2} = 0

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

2 u + 2 u^{2} = 0

Write in the standard form

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

2 u^{2} + 2 u = 0

Solve with the quadratic formula

2 u^{2} + 2 u = 0

Quadratic Equation Formula:

For

a = 2, b = 2, c = 0

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

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

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

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

Apply rule

0 \cdot a = 0

= 2^{2} - 0

2^{2} - 0 = 2^{2}

= 2^{2}

Apply radical rule:

assuming

a \geq 0

= 2

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

Separate the solutions

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

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

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

Add/Subtract the numbers:

- 2 + 2 = 0

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

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

= 0

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

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

Subtract the numbers:

- 2 - 2 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{4}

Apply the fraction rule:

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

= - \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 0, u = - 1

Substitute back

u = cos (θ)

cos (θ) = 0, cos (θ) = - 1

cos (θ) = 0, cos (θ) = - 1

cos (θ) = 0 : θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = 0

General solutions for

cos (θ) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

General solutions for

cos (θ) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = π + 2 πn

θ = π + 2 πn

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = π + 2 πn

Verify solutions by plugging them into the original equation

Check the solutions by plugging them into

- csc (θ) + 5 = cot (θ) + 6

Remove the ones that don't agree with the equation.

Check the solution

\frac{π}{2} + 2 πn :

False

\frac{π}{2} + 2 πn

Plug in

n = 1

\frac{π}{2} + 2 π 1

For

- csc (θ) + 5 = cot (θ) + 6

plug in

θ = \frac{π}{2} + 2 π 1

- csc (\frac{π}{2} + 2 π 1) + 5 = cot (\frac{π}{2} + 2 π 1) + 6

Refine

4 = 6

\Rightarrow False

Check the solution

\frac{3 π}{2} + 2 πn :

True

\frac{3 π}{2} + 2 πn

Plug in

n = 1

\frac{3 π}{2} + 2 π 1

For

- csc (θ) + 5 = cot (θ) + 6

plug in

θ = \frac{3 π}{2} + 2 π 1

- csc (\frac{3 π}{2} + 2 π 1) + 5 = cot (\frac{3 π}{2} + 2 π 1) + 6

Refine

6 = 6

\Rightarrow True

Check the solution

π + 2 πn :

True

π + 2 πn

Plug in

n = 1

π + 2 π 1

For

- csc (θ) + 5 = cot (θ) + 6

plug in

θ = π + 2 π 1

- csc (π + 2 π 1) + 5 = cot (π + 2 π 1) + 6

Refine

- \infty = - \infty

\Rightarrow True

θ = \frac{3 π}{2} + 2 πn, θ = π + 2 πn

Since the equation is undefined for:

π + 2 πn

θ = \frac{3 π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for -csc(θ)+5=cot(θ)+6 ?
The general solution for -csc(θ)+5=cot(θ)+6 is θ=(3pi)/2+2pin