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

The general solution for tan(x)+sqrt(3)=1+sqrt(3)cot(x) is x=(2pi)/3+pin,x= pi/4+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

x = \frac{2 π}{3} + πn, x = \frac{π}{4} + πn

Degrees

x = 12 0^{\circ} + 18 0^{\circ} n, x = 4 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Subtract

1 + 3 cot (x)

from both sides

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

Solve by substitution

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

Let:

cot (x) = u

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

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

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

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

Simplify

- 1 \cdot u : - u

- 1 \cdot u

Multiply:

1 \cdot u = u

= - 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

- u 3 u : - 3 u^{2}

- u 3 u

Apply exponent rule:

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

uu = u^{1 + 1}

= - 3 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= - 3 u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

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

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

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

Solve

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

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

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

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

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

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

Apply rule

- (- a) = a

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

Multiply the numbers:

4 \cdot 1 = 4

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

Simplify

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

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

Remove parentheses:

(- a) = - a

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

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

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

n

is even

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

= 1^{2}

Apply rule

1^{a} = 1

= 1

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 ( - 3 )}

Separate the solutions

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

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

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

Remove parentheses:

(- a) = - a

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

Apply the fraction rule:

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

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

Expand

- (- 1 + 3) + 3 + 1 : 2

- (- 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 : 2

1 - 3 + 3 + 1

Add similar elements:

- 3 + 3 = 0

= 1 + 1

Add the numbers:

1 + 1 = 2

= 2

= 2

= - \frac{2}{2 3}

Divide the numbers:

\frac{2}{2} = 1

= - \frac{1}{3}

Rationalize

- \frac{1}{3} : - \frac{3}{3}

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= - \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

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

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

Remove parentheses:

(- a) = - a

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

Apply the fraction rule:

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

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

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

Remove parentheses:

(a) = a

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

1 + 3 + 3 - 1 = 23

1 + 3 + 3 - 1

Add similar elements:

3 + 3 = 23

= 1 + 23 - 1

1 - 1 = 0

= 23

= \frac{2 3}{2 3}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

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

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

Substitute back

u = cot (x)

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

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

cot (x) = - \frac{3}{3} : x = \frac{2 π}{3} + πn

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

General solutions for

cot (x) = - \frac{3}{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{2 π}{3} + πn

x = \frac{2 π}{3} + 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{2 π}{3} + πn, x = \frac{π}{4} + πn

Popular Examples

tan(u)-cot(u)=2

tan (u) - cot (u) = 2

sin(x)sin(x)=0

sin (x) sin (x) = 0

15=arctan((0.375)/(2x))

15 = arctan (\frac{0.375}{2 x})

cos(θ)(1+cos(θ))=sin^2(θ)

cos (θ) (1 + cos (θ)) = sin^{2} (θ)

cos(θ)=-0.8,sin(pi^2-θ)

cos (θ) = - 0.8, sin (π^{2} - θ)

Frequently Asked Questions (FAQ)

What is the general solution for tan(x)+sqrt(3)=1+sqrt(3)cot(x) ?
The general solution for tan(x)+sqrt(3)=1+sqrt(3)cot(x) is x=(2pi)/3+pin,x= pi/4+pin