What is the general solution for tan(θ)-7=-12cot(θ) ?

The general solution for tan(θ)-7=-12cot(θ) is θ=1.24904…+pin,θ=1.32581…+pin

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tan(θ)-7=-12cot(θ)

Solution

tan (θ) - 7 = - 12 cot (θ)

Solution

θ = 1.24904 \dots + πn, θ = 1.32581 \dots + πn

Degrees

θ = 71.56505 \dots^{\circ} + 18 0^{\circ} n, θ = 75.96375 \dots^{\circ} + 18 0^{\circ} n

Solution steps

tan (θ) - 7 = - 12 cot (θ)

Subtract

- 12 cot (θ)

from both sides

tan (θ) - 7 + 12 cot (θ) = 0

Rewrite using trig identities

- 7 + tan (θ) + 12 cot (θ)

Use the basic trigonometric identity:

tan (x) = \frac{1}{cot ( x )}

= - 7 + \frac{1}{cot ( θ )} + 12 cot (θ)

- 7 + \frac{1}{cot ( θ )} + 12 cot (θ) = 0

Solve by substitution

- 7 + \frac{1}{cot ( θ )} + 12 cot (θ) = 0

Let:

cot (θ) = u

- 7 + \frac{1}{u} + 12 u = 0

- 7 + \frac{1}{u} + 12 u = 0 : u = \frac{1}{3}, u = \frac{1}{4}

- 7 + \frac{1}{u} + 12 u = 0

Multiply both sides by

u

- 7 + \frac{1}{u} + 12 u = 0

Multiply both sides by

u

- 7 u + \frac{1}{u} u + 12 uu = 0 \cdot u

Simplify

- 7 u + \frac{1}{u} u + 12 uu = 0 \cdot u

Simplify

\frac{1}{u} u : 1

\frac{1}{u} u

Multiply fractions:

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

= \frac{1 \cdot u}{u}

Cancel the common factor:

u

= 1

Simplify

12 uu : 12 u^{2}

12 uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= 12 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 12 u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

- 7 u + 1 + 12 u^{2} = 0

- 7 u + 1 + 12 u^{2} = 0

- 7 u + 1 + 12 u^{2} = 0

Solve

- 7 u + 1 + 12 u^{2} = 0 : u = \frac{1}{3}, u = \frac{1}{4}

- 7 u + 1 + 12 u^{2} = 0

Write in the standard form

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

12 u^{2} - 7 u + 1 = 0

Solve with the quadratic formula

12 u^{2} - 7 u + 1 = 0

Quadratic Equation Formula:

For

a = 12, b = - 7, c = 1

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 12 \cdot 1}{2 \cdot 12}

u_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 \cdot 12 \cdot 1}{2 \cdot 12}

(- 7)^{2} - 4 \cdot 12 \cdot 1 = 1

(- 7)^{2} - 4 \cdot 12 \cdot 1

Apply exponent rule:

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

n

is even

(- 7)^{2} = 7^{2}

= 7^{2} - 4 \cdot 12 \cdot 1

Multiply the numbers:

4 \cdot 12 \cdot 1 = 48

= 7^{2} - 48

7^{2} = 49

= 49 - 48

Subtract the numbers:

49 - 48 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 7 ) \pm 1}{2 \cdot 12}

Separate the solutions

u_{1} = \frac{- ( - 7 ) + 1}{2 \cdot 12}, u_{2} = \frac{- ( - 7 ) - 1}{2 \cdot 12}

u = \frac{- ( - 7 ) + 1}{2 \cdot 12} : \frac{1}{3}

\frac{- ( - 7 ) + 1}{2 \cdot 12}

Apply rule

- (- a) = a

= \frac{7 + 1}{2 \cdot 12}

Add the numbers:

7 + 1 = 8

= \frac{8}{2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{8}{24}

Cancel the common factor:

8

= \frac{1}{3}

u = \frac{- ( - 7 ) - 1}{2 \cdot 12} : \frac{1}{4}

\frac{- ( - 7 ) - 1}{2 \cdot 12}

Apply rule

- (- a) = a

= \frac{7 - 1}{2 \cdot 12}

Subtract the numbers:

7 - 1 = 6

= \frac{6}{2 \cdot 12}

Multiply the numbers:

2 \cdot 12 = 24

= \frac{6}{24}

Cancel the common factor:

6

= \frac{1}{4}

The solutions to the quadratic equation are:

u = \frac{1}{3}, u = \frac{1}{4}

u = \frac{1}{3}, u = \frac{1}{4}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

- 7 + \frac{1}{u} + 12 u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{1}{3}, u = \frac{1}{4}

Substitute back

u = cot (θ)

cot (θ) = \frac{1}{3}, cot (θ) = \frac{1}{4}

cot (θ) = \frac{1}{3}, cot (θ) = \frac{1}{4}

cot (θ) = \frac{1}{3} : θ = arccot (\frac{1}{3}) + πn

cot (θ) = \frac{1}{3}

Apply trig inverse properties

cot (θ) = \frac{1}{3}

General solutions for

cot (θ) = \frac{1}{3}

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

θ = arccot (\frac{1}{3}) + πn

θ = arccot (\frac{1}{3}) + πn

cot (θ) = \frac{1}{4} : θ = arccot (\frac{1}{4}) + πn

cot (θ) = \frac{1}{4}

Apply trig inverse properties

cot (θ) = \frac{1}{4}

General solutions for

cot (θ) = \frac{1}{4}

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

θ = arccot (\frac{1}{4}) + πn

θ = arccot (\frac{1}{4}) + πn

Combine all the solutions

θ = arccot (\frac{1}{3}) + πn, θ = arccot (\frac{1}{4}) + πn

Show solutions in decimal form

θ = 1.24904 \dots + πn, θ = 1.32581 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan(θ)-7=-12cot(θ) ?
The general solution for tan(θ)-7=-12cot(θ) is θ=1.24904…+pin,θ=1.32581…+pin