What is the general solution for tan(θ)+cot(θ)= 4/(sqrt(3)) ?

The general solution for tan(θ)+cot(θ)= 4/(sqrt(3)) is θ= pi/6+pin,θ= pi/3+pin

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tan(θ)+cot(θ)= 4/(sqrt(3))

Solution

tan (θ) + cot (θ) = \frac{4}{3}

Solution

θ = \frac{π}{6} + πn, θ = \frac{π}{3} + πn

Degrees

θ = 3 0^{\circ} + 18 0^{\circ} n, θ = 6 0^{\circ} + 18 0^{\circ} n

Solution steps

tan (θ) + cot (θ) = \frac{4}{3}

Subtract

\frac{4}{3}

from both sides

tan (θ) + cot (θ) - \frac{4}{3} = 0

Simplify

tan (θ) + cot (θ) - \frac{4}{3} : \frac{3 tan ( θ ) + 3 cot ( θ ) - 4}{3}

tan (θ) + cot (θ) - \frac{4}{3}

Convert element to fraction:

tan (θ) = \frac{tan ( θ ) 3}{3}, cot (θ) = \frac{cot ( θ ) 3}{3}

= \frac{tan ( θ ) 3}{3} + \frac{cot ( θ ) 3}{3} - \frac{4}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{tan ( θ ) 3 + cot ( θ ) 3 - 4}{3}

\frac{3 tan ( θ ) + 3 cot ( θ ) - 4}{3} = 0

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

3 tan (θ) + 3 cot (θ) - 4 = 0

Rewrite using trig identities

- 4 + cot (θ) 3 + 3 tan (θ)

Use the basic trigonometric identity:

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

= - 4 + cot (θ) 3 + 3 \frac{1}{cot ( θ )}

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

3 \frac{1}{cot ( θ )}

Multiply fractions:

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

= \frac{1 \cdot 3}{cot ( θ )}

Multiply:

1 \cdot 3 = 3

= \frac{3}{cot ( θ )}

= - 4 + 3 cot (θ) + \frac{3}{cot ( θ )}

- 4 + \frac{3}{cot ( θ )} + cot (θ) 3 = 0

Solve by substitution

- 4 + \frac{3}{cot ( θ )} + cot (θ) 3 = 0

Let:

cot (θ) = u

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

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

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

Multiply both sides by

u

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

Multiply both sides by

u

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

Simplify

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

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

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

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

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

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

Solve

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

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

Write in the standard form

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

3 u^{2} - 4 u + 3 = 0

Solve with the quadratic formula

3 u^{2} - 4 u + 3 = 0

Quadratic Equation Formula:

For

a = 3, b = - 4, c = 3

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

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

(- 4)^{2} - 433 = 2

(- 4)^{2} - 433

(- 4)^{2} = 4^{2}

(- 4)^{2}

Apply exponent rule:

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

n

is even

(- 4)^{2} = 4^{2}

= 4^{2}

433 = 12

433

Apply radical rule:

a a = a

33 = 3

= 4 \cdot 3

Multiply the numbers:

4 \cdot 3 = 12

= 12

= 4^{2} - 12

4^{2} = 16

= 16 - 12

Subtract the numbers:

16 - 12 = 4

= 4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

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

Separate the solutions

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

u = \frac{- ( - 4 ) + 2}{2 3} : 3

\frac{- ( - 4 ) + 2}{2 3}

Apply rule

- (- a) = a

= \frac{4 + 2}{2 3}

Add the numbers:

4 + 2 = 6

= \frac{6}{2 3}

Divide the numbers:

\frac{6}{2} = 3

= \frac{3}{3}

Apply radical rule:

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

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

Apply exponent rule:

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

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

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

Subtract the numbers:

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

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

Apply radical rule:

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

= 3

u = \frac{- ( - 4 ) - 2}{2 3} : \frac{3}{3}

\frac{- ( - 4 ) - 2}{2 3}

Apply rule

- (- a) = a

= \frac{4 - 2}{2 3}

Subtract the numbers:

4 - 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}

The solutions to the quadratic equation are:

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

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

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

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

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

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

Substitute back

u = cot (θ)

cot (θ) = 3, cot (θ) = \frac{3}{3}

cot (θ) = 3, cot (θ) = \frac{3}{3}

cot (θ) = 3 : θ = \frac{π}{6} + πn

cot (θ) = 3

General solutions for

cot (θ) = 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

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

cot (θ) = \frac{3}{3} : θ = \frac{π}{3} + πn

cot (θ) = \frac{3}{3}

General solutions for

cot (θ) = \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

θ = \frac{π}{3} + πn

θ = \frac{π}{3} + πn

Combine all the solutions

θ = \frac{π}{6} + πn, θ = \frac{π}{3} + πn

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Frequently Asked Questions (FAQ)

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