What is the general solution for 6cot(x)=5-tan(x) ?

The general solution for 6cot(x)=5-tan(x) is x=1.10714…+pin,x=1.24904…+pin

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6cot(x)=5-tan(x)

Solution

6 cot (x) = 5 - tan (x)

Solution

x = 1.10714 \dots + πn, x = 1.24904 \dots + πn

Degrees

x = 63.43494 \dots^{\circ} + 18 0^{\circ} n, x = 71.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

6 cot (x) = 5 - tan (x)

Subtract

5 - tan (x)

from both sides

6 cot (x) - 5 + tan (x) = 0

Rewrite using trig identities

- 5 + tan (x) + 6 cot (x)

Use the basic trigonometric identity:

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

= - 5 + \frac{1}{cot ( x )} + 6 cot (x)

- 5 + \frac{1}{cot ( x )} + 6 cot (x) = 0

Solve by substitution

- 5 + \frac{1}{cot ( x )} + 6 cot (x) = 0

Let:

cot (x) = u

- 5 + \frac{1}{u} + 6 u = 0

- 5 + \frac{1}{u} + 6 u = 0 : u = \frac{1}{2}, u = \frac{1}{3}

- 5 + \frac{1}{u} + 6 u = 0

Multiply both sides by

u

- 5 + \frac{1}{u} + 6 u = 0

Multiply both sides by

u

- 5 u + \frac{1}{u} u + 6 uu = 0 \cdot u

Simplify

- 5 u + \frac{1}{u} u + 6 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

6 uu : 6 u^{2}

6 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= 6 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= 6 u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

- 5 u + 1 + 6 u^{2} = 0

- 5 u + 1 + 6 u^{2} = 0

- 5 u + 1 + 6 u^{2} = 0

Solve

- 5 u + 1 + 6 u^{2} = 0 : u = \frac{1}{2}, u = \frac{1}{3}

- 5 u + 1 + 6 u^{2} = 0

Write in the standard form

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

6 u^{2} - 5 u + 1 = 0

Solve with the quadratic formula

6 u^{2} - 5 u + 1 = 0

Quadratic Equation Formula:

For

a = 6, b = - 5, c = 1

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

u_{1, 2} = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 \cdot 6 \cdot 1}{2 \cdot 6}

(- 5)^{2} - 4 \cdot 6 \cdot 1 = 1

(- 5)^{2} - 4 \cdot 6 \cdot 1

Apply exponent rule:

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

n

is even

(- 5)^{2} = 5^{2}

= 5^{2} - 4 \cdot 6 \cdot 1

Multiply the numbers:

4 \cdot 6 \cdot 1 = 24

= 5^{2} - 24

5^{2} = 25

= 25 - 24

Subtract the numbers:

25 - 24 = 1

= 1

Apply rule

1 = 1

= 1

u_{1, 2} = \frac{- ( - 5 ) \pm 1}{2 \cdot 6}

Separate the solutions

u_{1} = \frac{- ( - 5 ) + 1}{2 \cdot 6}, u_{2} = \frac{- ( - 5 ) - 1}{2 \cdot 6}

u = \frac{- ( - 5 ) + 1}{2 \cdot 6} : \frac{1}{2}

\frac{- ( - 5 ) + 1}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{5 + 1}{2 \cdot 6}

Add the numbers:

5 + 1 = 6

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

Multiply the numbers:

2 \cdot 6 = 12

= \frac{6}{12}

Cancel the common factor:

6

= \frac{1}{2}

u = \frac{- ( - 5 ) - 1}{2 \cdot 6} : \frac{1}{3}

\frac{- ( - 5 ) - 1}{2 \cdot 6}

Apply rule

- (- a) = a

= \frac{5 - 1}{2 \cdot 6}

Subtract the numbers:

5 - 1 = 4

= \frac{4}{2 \cdot 6}

Multiply the numbers:

2 \cdot 6 = 12

= \frac{4}{12}

Cancel the common factor:

4

= \frac{1}{3}

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = \frac{1}{3}

u = \frac{1}{2}, u = \frac{1}{3}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

- 5 + \frac{1}{u} + 6 u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{1}{2}, u = \frac{1}{3}

Substitute back

u = cot (x)

cot (x) = \frac{1}{2}, cot (x) = \frac{1}{3}

cot (x) = \frac{1}{2}, cot (x) = \frac{1}{3}

cot (x) = \frac{1}{2} : x = arccot (\frac{1}{2}) + πn

cot (x) = \frac{1}{2}

Apply trig inverse properties

cot (x) = \frac{1}{2}

General solutions for

cot (x) = \frac{1}{2}

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

x = arccot (\frac{1}{2}) + πn

x = arccot (\frac{1}{2}) + πn

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

cot (x) = \frac{1}{3}

Apply trig inverse properties

cot (x) = \frac{1}{3}

General solutions for

cot (x) = \frac{1}{3}

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

x = arccot (\frac{1}{3}) + πn

x = arccot (\frac{1}{3}) + πn

Combine all the solutions

x = arccot (\frac{1}{2}) + πn, x = arccot (\frac{1}{3}) + πn

Show solutions in decimal form

x = 1.10714 \dots + πn, x = 1.24904 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 6cot(x)=5-tan(x) ?
The general solution for 6cot(x)=5-tan(x) is x=1.10714…+pin,x=1.24904…+pin