What is the general solution for 9tan(2x)-9cot(x)=0 ?

The general solution for 9tan(2x)-9cot(x)=0 is x=0.52359…+pin,x=-0.52359…+pin

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9tan(2x)-9cot(x)=0

Solution

9 tan (2 x) - 9 cot (x) = 0

Solution

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

Degrees

x = 3 0^{\circ} + 18 0^{\circ} n, x = - 3 0^{\circ} + 18 0^{\circ} n

Solution steps

9 tan (2 x) - 9 cot (x) = 0

Rewrite using trig identities

- 9 cot (x) + 9 tan (2 x)

tan (2 x) = \frac{2 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

tan (2 x)

Use the Double Angle identity:

tan (2 x) = \frac{2 tan ( x )}{1 - tan ^{2} ( x )}

= \frac{2 tan ( x )}{1 - tan ^{2} ( x )}

Factor

\frac{2 tan ( x )}{1 - tan ^{2} ( x )} : \frac{2 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

\frac{2 tan ( x )}{1 - tan ^{2} ( x )}

Factor

1 - tan^{2} (x) : (1 + tan (x)) (1 - tan (x))

1 - tan^{2} (x)

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

1 - tan^{2} (x) = (1 + tan (x)) (1 - tan (x))

= (1 + tan (x)) (1 - tan (x))

= \frac{2 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

= \frac{2 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

= - 9 cot (x) + 9 \cdot \frac{2 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

9 \cdot \frac{2 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )} = \frac{18 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

9 \cdot \frac{2 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

Multiply fractions:

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

= \frac{2 tan ( x ) \cdot 9}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

Multiply the numbers:

2 \cdot 9 = 18

= \frac{18 tan ( x )}{( tan ( x ) + 1 ) ( - tan ( x ) + 1 )}

= - 9 cot (x) + \frac{18 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )}

Use the basic trigonometric identity:

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

= \frac{18 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )} - 9 \cdot \frac{1}{tan ( x )}

Simplify

\frac{18 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )} - 9 \cdot \frac{1}{tan ( x )} : - \frac{27 tan ^{2} ( x ) - 9}{tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

\frac{18 tan ( x )}{( 1 + tan ( x ) ) ( 1 - tan ( x ) )} - 9 \cdot \frac{1}{tan ( x )}

9 \cdot \frac{1}{tan ( x )} = \frac{9}{tan ( x )}

9 \cdot \frac{1}{tan ( x )}

Multiply fractions:

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

= \frac{1 \cdot 9}{tan ( x )}

Multiply the numbers:

1 \cdot 9 = 9

= \frac{9}{tan ( x )}

= \frac{18 tan ( x )}{( tan ( x ) + 1 ) ( - tan ( x ) + 1 )} - \frac{9}{tan ( x )}

Factor

(1 + tan (x)) (1 - tan (x)) : - (1 + tan (x)) (tan (x) - 1)

(1 + tan (x)) (1 - tan (x))

Factor

1 - tan (x) : - (tan (x) - 1)

1 - tan (x)

Factor out common term

- 1

= - (tan (x) - 1)

= - (1 + tan (x)) (tan (x) - 1)

= \frac{18 tan ( x )}{- ( 1 + tan ( x ) ) ( tan ( x ) - 1 )} - \frac{9}{tan ( x )}

Least Common Multiplier of

- (1 + tan (x)) (tan (x) - 1), tan (x) : - tan (x) (tan (x) + 1) (tan (x) - 1)

- (1 + tan (x)) (tan (x) - 1), tan (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

- (1 + tan (x)) (tan (x) - 1)

tan (x)

= - tan (x) (tan (x) + 1) (tan (x) - 1)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

- tan (x) (tan (x) + 1) (tan (x) - 1)

For

\frac{18 tan ( x )}{- ( 1 + tan ( x ) ) ( tan ( x ) - 1 )} :

multiply the denominator and numerator by

tan (x)

\frac{18 tan ( x )}{- ( 1 + tan ( x ) ) ( tan ( x ) - 1 )} = \frac{18 tan ( x ) tan ( x )}{( - ( 1 + tan ( x ) ) ( tan ( x ) - 1 ) ) tan ( x )} = \frac{18 tan ^{2} ( x )}{- tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

For

\frac{9}{tan ( x )} :

multiply the denominator and numerator by

- (tan (x) + 1) (tan (x) - 1)

\frac{9}{tan ( x )} = \frac{9 ( - ( tan ( x ) + 1 ) ( tan ( x ) - 1 ) )}{tan ( x ) ( - ( tan ( x ) + 1 ) ( tan ( x ) - 1 ) )} = \frac{- 9 ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}{- tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

= \frac{18 tan ^{2} ( x )}{- tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )} - \frac{- 9 ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}{- tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

Since the denominators are equal, combine the fractions:

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

= \frac{18 tan ^{2} ( x ) - ( - 9 ( tan ( x ) + 1 ) ( tan ( x ) - 1 ) )}{- tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

Refine

= - \frac{18 tan ^{2} ( x ) + 9 ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}{tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

Expand

18 tan^{2} (x) + 9 (tan (x) + 1) (tan (x) - 1) : 27 tan^{2} (x) - 9

18 tan^{2} (x) + 9 (tan (x) + 1) (tan (x) - 1)

Expand

9 (tan (x) + 1) (tan (x) - 1) : 9 tan^{2} (x) - 9

Expand

(tan (x) + 1) (tan (x) - 1) : tan^{2} (x) - 1

(tan (x) + 1) (tan (x) - 1)

Apply Difference of Two Squares Formula:

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

a = tan (x), b = 1

= tan^{2} (x) - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= tan^{2} (x) - 1

= 9 (tan^{2} (x) - 1)

Expand

9 (tan^{2} (x) - 1) : 9 tan^{2} (x) - 9

9 (tan^{2} (x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = 9, b = tan^{2} (x), c = 1

= 9 tan^{2} (x) - 9 \cdot 1

Multiply the numbers:

9 \cdot 1 = 9

= 9 tan^{2} (x) - 9

= 9 tan^{2} (x) - 9

= 18 tan^{2} (x) + 9 tan^{2} (x) - 9

Add similar elements:

18 tan^{2} (x) + 9 tan^{2} (x) = 27 tan^{2} (x)

= 27 tan^{2} (x) - 9

= - \frac{27 tan ^{2} ( x ) - 9}{tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

= - \frac{27 tan ^{2} ( x ) - 9}{tan ( x ) ( tan ( x ) + 1 ) ( tan ( x ) - 1 )}

- \frac{- 9 + 27 tan ^{2} ( x )}{( - 1 + tan ( x ) ) ( 1 + tan ( x ) ) tan ( x )} = 0

Solve by substitution

- \frac{- 9 + 27 tan ^{2} ( x )}{( - 1 + tan ( x ) ) ( 1 + tan ( x ) ) tan ( x )} = 0

Let:

tan (x) = u

- \frac{- 9 + 27 u ^{2}}{( - 1 + u ) ( 1 + u ) u} = 0

- \frac{- 9 + 27 u ^{2}}{( - 1 + u ) ( 1 + u ) u} = 0 : u = \frac{1}{3}, u = - \frac{1}{3}

- \frac{- 9 + 27 u ^{2}}{( - 1 + u ) ( 1 + u ) u} = 0

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

- 9 + 27 u^{2} = 0

Solve

- 9 + 27 u^{2} = 0 : u = \frac{1}{3}, u = - \frac{1}{3}

- 9 + 27 u^{2} = 0

Move

9

to the right side

- 9 + 27 u^{2} = 0

Add

9

to both sides

- 9 + 27 u^{2} + 9 = 0 + 9

Simplify

27 u^{2} = 9

27 u^{2} = 9

Divide both sides by

27

27 u^{2} = 9

Divide both sides by

27

\frac{27 u ^{2}}{27} = \frac{9}{27}

Simplify

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

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

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

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

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

Verify Solutions

Find undefined (singularity) points:

u = 1, u = - 1, u = 0

Take the denominator(s) of

- \frac{- 9 + 27 u ^{2}}{( - 1 + u ) ( 1 + u ) u}

and compare to zero

Solve

(- 1 + u) (1 + u) u = 0 : u = 1, u = - 1, u = 0

(- 1 + u) (1 + u) u = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

- 1 + u = 0 or 1 + u = 0 or u = 0

Solve

- 1 + u = 0 : u = 1

- 1 + u = 0

Move

1

to the right side

- 1 + u = 0

Add

1

to both sides

- 1 + u + 1 = 0 + 1

Simplify

u = 1

u = 1

Solve

1 + u = 0 : u = - 1

1 + u = 0

Move

1

to the right side

1 + u = 0

Subtract

1

from both sides

1 + u - 1 = 0 - 1

Simplify

u = - 1

u = - 1

The solutions are

u = 1, u = - 1, u = 0

The following points are undefined

u = 1, u = - 1, u = 0

Combine undefined points with solutions:

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

Substitute back

u = tan (x)

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{3}

tan (x) = \frac{1}{3}, tan (x) = - \frac{1}{3}

tan (x) = \frac{1}{3} : x = arctan (\frac{1}{3}) + πn

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

Apply trig inverse properties

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

General solutions for

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

tan (x) = a \Rightarrow x = arctan (a) + πn

x = arctan (\frac{1}{3}) + πn

x = arctan (\frac{1}{3}) + πn

tan (x) = - \frac{1}{3} : x = arctan (- \frac{1}{3}) + πn

tan (x) = - \frac{1}{3}

Apply trig inverse properties

tan (x) = - \frac{1}{3}

General solutions for

tan (x) = - \frac{1}{3}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{1}{3}) + πn

x = arctan (- \frac{1}{3}) + πn

Combine all the solutions

x = arctan (\frac{1}{3}) + πn, x = arctan (- \frac{1}{3}) + πn

Show solutions in decimal form

x = 0.52359 \dots + πn, x = - 0.52359 \dots + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 9tan(2x)-9cot(x)=0 ?
The general solution for 9tan(2x)-9cot(x)=0 is x=0.52359…+pin,x=-0.52359…+pin