What is the general solution for tan(x)-1=-1/(sqrt(3))+1/(sqrt(3))cot(x) ?

The general solution for tan(x)-1=-1/(sqrt(3))+1/(sqrt(3))cot(x) is x=(5pi)/6+pin,x= pi/4+pin

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tan(x)-1=-1/(sqrt(3))+1/(sqrt(3))cot(x)

Solution

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

Solution

x = \frac{5 π}{6} + πn, x = \frac{π}{4} + πn

Degrees

x = 15 0^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

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

from both sides

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

Simplify

tan (x) - 1 - \frac{- 1 + cot ( x )}{3} : \frac{3 tan ( x ) - 3 + 1 - cot ( x )}{3}

tan (x) - 1 - \frac{- 1 + cot ( x )}{3}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot 3 = 3

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

Expand

tan (x) 3 - 3 - (- 1 + cot (x)) : tan (x) 3 - 3 + 1 - cot (x)

tan (x) 3 - 3 - (- 1 + cot (x))

= 3 tan (x) - 3 - (- 1 + cot (x))

- (- 1 + cot (x)) : 1 - cot (x)

- (- 1 + cot (x))

Distribute parentheses

= - (- 1) - (cot (x))

Apply minus-plus rules

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

= 1 - cot (x)

= tan (x) 3 - 3 + 1 - cot (x)

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

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

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

3 tan (x) - 3 + 1 - cot (x) = 0

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Multiply:

1 \cdot 3 = 3

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

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

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

Solve by substitution

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

Let:

cot (x) = u

1 - u + \frac{3}{u} - 3 = 0

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

1 - u + \frac{3}{u} - 3 = 0

Multiply both sides by

u

1 - u + \frac{3}{u} - 3 = 0

Multiply both sides by

u

1 \cdot u - uu + \frac{3}{u} u - 3 u = 0 \cdot u

Simplify

1 \cdot u - uu + \frac{3}{u} u - 3 u = 0 \cdot u

Simplify

1 \cdot u : u

1 \cdot u

Multiply:

1 \cdot u = u

= 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{3}{u} u : 3

\frac{3}{u} u

Multiply fractions:

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

= \frac{3 u}{u}

Cancel the common factor:

u

= 3

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

u - u^{2} + 3 - 3 u = 0

u - u^{2} + 3 - 3 u = 0

u - u^{2} + 3 - 3 u = 0

Solve

u - u^{2} + 3 - 3 u = 0 : u = - 3, u = 1

u - u^{2} + 3 - 3 u = 0

Write in the standard form

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

- u^{2} + (1 - 3) u + 3 = 0

Solve with the quadratic formula

- u^{2} + (1 - 3) u + 3 = 0

Quadratic Equation Formula:

For

a = - 1, b = 1 - 3, c = 3

u_{1, 2} = \frac{- ( 1 - 3 ) \pm ( 1 - 3 ) ^{2} - 4 ( - 1 ) 3}{2 ( - 1 )}

u_{1, 2} = \frac{- ( 1 - 3 ) \pm ( 1 - 3 ) ^{2} - 4 ( - 1 ) 3}{2 ( - 1 )}

(1 - 3)^{2} - 4 (- 1) 3 = 3 + 1

(1 - 3)^{2} - 4 (- 1) 3

Apply rule

- (- a) = a

= (1 - 3)^{2} + 4 \cdot 1 \cdot 3

Multiply the numbers:

4 \cdot 1 = 4

= (1 - 3)^{2} + 43

Expand

(1 - 3)^{2} + 43 : 4 + 23

(1 - 3)^{2} + 43

(1 - 3)^{2} : 4 - 23

Apply Perfect Square Formula:

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

a = 1, b = 3

= 1^{2} - 2 \cdot 1 \cdot 3 + (3)^{2}

Simplify

1^{2} - 2 \cdot 1 \cdot 3 + (3)^{2} : 4 - 23

1^{2} - 2 \cdot 1 \cdot 3 + (3)^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 2 \cdot 1 \cdot 3 + (3)^{2}

2 \cdot 1 \cdot 3 = 23

2 \cdot 1 \cdot 3

Multiply the numbers:

2 \cdot 1 = 2

= 23

(3)^{2} = 3

(3)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 3

= 1 - 23 + 3

Add the numbers:

1 + 3 = 4

= 4 - 23

= 4 - 23

= 4 - 23 + 43

Add similar elements:

- 23 + 43 = 23

= 4 + 23

= 4 + 23

= 3 + 23 + 1

= (3)^{2} + 23 + (1)^{2}

1 = 1

1

Apply rule

1 = 1

= 1

= (3)^{2} + 23 + 1^{2}

23 \cdot 1 = 23

23 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 23

= (3)^{2} + 23 \cdot 1 + 1^{2}

Apply Perfect Square Formula:

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

(3)^{2} + 23 \cdot 1 + 1^{2} = (3 + 1)^{2}

= (3 + 1)^{2}

Apply radical rule:

(3 + 1)^{2} = 3 + 1

= 3 + 1

u_{1, 2} = \frac{- ( 1 - 3 ) \pm ( 3 + 1 )}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- ( 1 - 3 ) + 3 + 1}{2 ( - 1 )}, u_{2} = \frac{- ( 1 - 3 ) - ( 3 + 1 )}{2 ( - 1 )}

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

\frac{- ( 1 - 3 ) + 3 + 1}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- ( 1 - 3 ) + 3 + 1}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- ( 1 - 3 ) + 3 + 1}{- 2}

Apply the fraction rule:

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

= - \frac{- ( 1 - 3 ) + 3 + 1}{2}

Expand

- (1 - 3) + 3 + 1 : 23

- (1 - 3) + 3 + 1

- (1 - 3) : - 1 + 3

- (1 - 3)

Distribute parentheses

= - (1) - (- 3)

Apply minus-plus rules

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

= - 1 + 3

= - 1 + 3 + 3 + 1

Simplify

- 1 + 3 + 3 + 1 : 23

- 1 + 3 + 3 + 1

Add similar elements:

3 + 3 = 23

= - 1 + 23 + 1

- 1 + 1 = 0

= 23

= 23

= - \frac{2 3}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 3

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

\frac{- ( 1 - 3 ) - ( 3 + 1 )}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- ( 1 - 3 ) - ( 3 + 1 )}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- ( 1 - 3 ) - ( 1 + 3 )}{- 2}

Apply the fraction rule:

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

- (1 - 3) - (3 + 1) = - ((1 - 3) + (1 + 3))

= \frac{( 1 - 3 ) + ( 1 + 3 )}{2}

Remove parentheses:

(a) = a

= \frac{1 - 3 + 1 + 3}{2}

1 - 3 + 1 + 3 = 2

1 - 3 + 1 + 3

Add similar elements:

- 3 + 3 = 0

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

= \frac{2}{2}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - 3, u = 1

u = - 3, u = 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

1 - u + \frac{3}{u} - 3

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 3, u = 1

Substitute back

u = cot (x)

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

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

cot (x) = - 3 : x = \frac{5 π}{6} + πn

cot (x) = - 3

General solutions for

cot (x) = - 3

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

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

cot (x) = 1 : x = \frac{π}{4} + πn

cot (x) = 1

General solutions for

cot (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

x = \frac{π}{4} + πn

x = \frac{π}{4} + πn

Combine all the solutions

x = \frac{5 π}{6} + πn, x = \frac{π}{4} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for tan(x)-1=-1/(sqrt(3))+1/(sqrt(3))cot(x) ?
The general solution for tan(x)-1=-1/(sqrt(3))+1/(sqrt(3))cot(x) is x=(5pi)/6+pin,x= pi/4+pin