What is the general solution for cot^2(3x)-tan^2(3x)=0 ?

The general solution for cot^2(3x)-tan^2(3x)=0 is x= pi/(12)+(pin)/3 ,x= pi/4+(pin)/3

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cot^2(3x)-tan^2(3x)=0

Solution

cot^{2} (3 x) - tan^{2} (3 x) = 0

Solution

x = \frac{π}{12} + \frac{πn}{3}, x = \frac{π}{4} + \frac{πn}{3}

Degrees

x = 1 5^{\circ} + 6 0^{\circ} n, x = 4 5^{\circ} + 6 0^{\circ} n

Solution steps

cot^{2} (3 x) - tan^{2} (3 x) = 0

Factor

cot^{2} (3 x) - tan^{2} (3 x) : (cot (3 x) + tan (3 x)) (cot (3 x) - tan (3 x))

cot^{2} (3 x) - tan^{2} (3 x)

Apply Difference of Two Squares Formula:

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

cot^{2} (3 x) - tan^{2} (3 x) = (cot (3 x) + tan (3 x)) (cot (3 x) - tan (3 x))

= (cot (3 x) + tan (3 x)) (cot (3 x) - tan (3 x))

(cot (3 x) + tan (3 x)) (cot (3 x) - tan (3 x)) = 0

Solving each part separately

cot (3 x) + tan (3 x) = 0 or cot (3 x) - tan (3 x) = 0

cot (3 x) + tan (3 x) = 0 :

No Solution

cot (3 x) + tan (3 x) = 0

Rewrite using trig identities

cot (3 x) + tan (3 x)

Use the basic trigonometric identity:

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

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

cot (3 x) + \frac{1}{cot ( 3 x )} = 0

Solve by substitution

cot (3 x) + \frac{1}{cot ( 3 x )} = 0

Let:

cot (3 x) = u

u + \frac{1}{u} = 0

u + \frac{1}{u} = 0 : u = i, u = - i

u + \frac{1}{u} = 0

Multiply both sides by

u

u + \frac{1}{u} = 0

Multiply both sides by

u

uu + \frac{1}{u} u = 0 \cdot u

Simplify

uu + \frac{1}{u} u = 0 \cdot u

Simplify

uu : u^{2}

uu

Apply exponent rule:

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

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

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

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

u^{2} + 1 = 0

u^{2} + 1 = 0

u^{2} + 1 = 0

Solve

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

u^{2} + 1 = 0

Move

1

to the right side

u^{2} + 1 = 0

Subtract

1

from both sides

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

Simplify

u^{2} = - 1

u^{2} = - 1

For

x^{2} = f (a)

the solutions are

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

u = - 1, u = - - 1

Simplify

- 1 : i

- 1

Apply imaginary number rule:

- 1 = i

= i

Simplify

- - 1 : - i

- - 1

Apply imaginary number rule:

- 1 = i

= - i

u = i, u = - i

u = i, u = - i

Substitute back

u = cot (3 x)

cot (3 x) = i, cot (3 x) = - i

cot (3 x) = i, cot (3 x) = - i

cot (3 x) = i :

No Solution

cot (3 x) = i

No Solution

cot (3 x) = - i :

No Solution

cot (3 x) = - i

No Solution

Combine all the solutions

No Solution

cot (3 x) - tan (3 x) = 0 : x = \frac{π}{12} + \frac{πn}{3}, x = \frac{π}{4} + \frac{πn}{3}

cot (3 x) - tan (3 x) = 0

Rewrite using trig identities

cot (3 x) - tan (3 x)

Use the basic trigonometric identity:

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

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

cot (3 x) - \frac{1}{cot ( 3 x )} = 0

Solve by substitution

cot (3 x) - \frac{1}{cot ( 3 x )} = 0

Let:

cot (3 x) = u

u - \frac{1}{u} = 0

u - \frac{1}{u} = 0 : u = 1, u = - 1

u - \frac{1}{u} = 0

Multiply both sides by

u

u - \frac{1}{u} = 0

Multiply both sides by

u

uu - \frac{1}{u} u = 0 \cdot u

Simplify

uu - \frac{1}{u} u = 0 \cdot u

Simplify

uu : u^{2}

uu

Apply exponent rule:

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

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

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

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

u^{2} - 1 = 0

u^{2} - 1 = 0

u^{2} - 1 = 0

Solve

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

u^{2} - 1 = 0

Move

1

to the right side

u^{2} - 1 = 0

Add

1

to both sides

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

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

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

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

u = 1, u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

u - \frac{1}{u}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 1, u = - 1

Substitute back

u = cot (3 x)

cot (3 x) = 1, cot (3 x) = - 1

cot (3 x) = 1, cot (3 x) = - 1

cot (3 x) = 1 : x = \frac{π}{12} + \frac{πn}{3}

cot (3 x) = 1

General solutions for

cot (3 x) = 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

3 x = \frac{π}{4} + πn

3 x = \frac{π}{4} + πn

Solve

3 x = \frac{π}{4} + πn : x = \frac{π}{12} + \frac{πn}{3}

3 x = \frac{π}{4} + πn

Divide both sides by

3

3 x = \frac{π}{4} + πn

Divide both sides by

3

\frac{3 x}{3} = \frac{\frac{π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 x}{3} = \frac{\frac{π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{π}{4}}{3} + \frac{πn}{3} : \frac{π}{12} + \frac{πn}{3}

\frac{\frac{π}{4}}{3} + \frac{πn}{3}

\frac{\frac{π}{4}}{3} = \frac{π}{12}

\frac{\frac{π}{4}}{3}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{π}{12}

= \frac{π}{12} + \frac{πn}{3}

x = \frac{π}{12} + \frac{πn}{3}

x = \frac{π}{12} + \frac{πn}{3}

x = \frac{π}{12} + \frac{πn}{3}

x = \frac{π}{12} + \frac{πn}{3}

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

cot (3 x) = - 1

General solutions for

cot (3 x) = - 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

3 x = \frac{3 π}{4} + πn

3 x = \frac{3 π}{4} + πn

Solve

3 x = \frac{3 π}{4} + πn : x = \frac{π}{4} + \frac{πn}{3}

3 x = \frac{3 π}{4} + πn

Divide both sides by

3

3 x = \frac{3 π}{4} + πn

Divide both sides by

3

\frac{3 x}{3} = \frac{\frac{3 π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 x}{3} = \frac{\frac{3 π}{4}}{3} + \frac{πn}{3}

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{3 π}{4}}{3} + \frac{πn}{3} : \frac{π}{4} + \frac{πn}{3}

\frac{\frac{3 π}{4}}{3} + \frac{πn}{3}

\frac{\frac{3 π}{4}}{3} = \frac{π}{4}

\frac{\frac{3 π}{4}}{3}

Apply the fraction rule:

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

= \frac{3 π}{4 \cdot 3}

Multiply the numbers:

4 \cdot 3 = 12

= \frac{3 π}{12}

Cancel the common factor:

3

= \frac{π}{4}

= \frac{π}{4} + \frac{πn}{3}

x = \frac{π}{4} + \frac{πn}{3}

x = \frac{π}{4} + \frac{πn}{3}

x = \frac{π}{4} + \frac{πn}{3}

x = \frac{π}{4} + \frac{πn}{3}

Combine all the solutions

x = \frac{π}{12} + \frac{πn}{3}, x = \frac{π}{4} + \frac{πn}{3}

Combine all the solutions

x = \frac{π}{12} + \frac{πn}{3}, x = \frac{π}{4} + \frac{πn}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for cot^2(3x)-tan^2(3x)=0 ?
The general solution for cot^2(3x)-tan^2(3x)=0 is x= pi/(12)+(pin)/3 ,x= pi/4+(pin)/3