What is the general solution for 6tan(θ)-2cot(θ)=0 ?

The general solution for 6tan(θ)-2cot(θ)=0 is θ= pi/6+pin,θ=(5pi)/6+pin

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Popular >

6tan(θ)-2cot(θ)=0

Solution

6 tan (θ) - 2 cot (θ) = 0

Solution

θ = \frac{π}{6} + πn, θ = \frac{5 π}{6} + πn

Degrees

θ = 3 0^{\circ} + 18 0^{\circ} n, θ = 15 0^{\circ} + 18 0^{\circ} n

Solution steps

6 tan (θ) - 2 cot (θ) = 0

Rewrite using trig identities

- 2 cot (θ) + 6 tan (θ)

Use the basic trigonometric identity:

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

= - 2 cot (θ) + 6 \cdot \frac{1}{cot ( θ )}

6 \cdot \frac{1}{cot ( θ )} = \frac{6}{cot ( θ )}

6 \cdot \frac{1}{cot ( θ )}

Multiply fractions:

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

= \frac{1 \cdot 6}{cot ( θ )}

Multiply the numbers:

1 \cdot 6 = 6

= \frac{6}{cot ( θ )}

= - 2 cot (θ) + \frac{6}{cot ( θ )}

\frac{6}{cot ( θ )} - 2 cot (θ) = 0

Solve by substitution

\frac{6}{cot ( θ )} - 2 cot (θ) = 0

Let:

cot (θ) = u

\frac{6}{u} - 2 u = 0

\frac{6}{u} - 2 u = 0 : u = 3, u = - 3

\frac{6}{u} - 2 u = 0

Multiply both sides by

u

\frac{6}{u} - 2 u = 0

Multiply both sides by

u

\frac{6}{u} u - 2 uu = 0 \cdot u

Simplify

\frac{6}{u} u - 2 uu = 0 \cdot u

Simplify

\frac{6}{u} u : 6

\frac{6}{u} u

Multiply fractions:

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

= \frac{6 u}{u}

Cancel the common factor:

u

= 6

Simplify

- 2 uu : - 2 u^{2}

- 2 uu

Apply exponent rule:

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

uu = u^{1 + 1}

= - 2 u^{1 + 1}

Add the numbers:

1 + 1 = 2

= - 2 u^{2}

Simplify

0 \cdot u : 0

0 \cdot u

Apply rule

0 \cdot a = 0

= 0

6 - 2 u^{2} = 0

6 - 2 u^{2} = 0

6 - 2 u^{2} = 0

Solve

6 - 2 u^{2} = 0 : u = 3, u = - 3

6 - 2 u^{2} = 0

Move

6

to the right side

6 - 2 u^{2} = 0

Subtract

6

from both sides

6 - 2 u^{2} - 6 = 0 - 6

Simplify

- 2 u^{2} = - 6

- 2 u^{2} = - 6

Divide both sides by

- 2

- 2 u^{2} = - 6

Divide both sides by

- 2

\frac{- 2 u ^{2}}{- 2} = \frac{- 6}{- 2}

Simplify

\frac{- 2 u ^{2}}{- 2} = \frac{- 6}{- 2}

Simplify

\frac{- 2 u ^{2}}{- 2} : u^{2}

\frac{- 2 u ^{2}}{- 2}

Apply the fraction rule:

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

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

Divide the numbers:

\frac{2}{2} = 1

= u^{2}

Simplify

\frac{- 6}{- 2} : 3

\frac{- 6}{- 2}

Apply the fraction rule:

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

= \frac{6}{2}

Divide the numbers:

\frac{6}{2} = 3

= 3

u^{2} = 3

u^{2} = 3

u^{2} = 3

For

x^{2} = f (a)

the solutions are

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

u = 3, u = - 3

u = 3, u = - 3

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

\frac{6}{u} - 2 u

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = 3, u = - 3

Substitute back

u = cot (θ)

cot (θ) = 3, cot (θ) = - 3

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

θ = \frac{5 π}{6} + πn

Combine all the solutions

θ = \frac{π}{6} + πn, θ = \frac{5 π}{6} + πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for 6tan(θ)-2cot(θ)=0 ?
The general solution for 6tan(θ)-2cot(θ)=0 is θ= pi/6+pin,θ=(5pi)/6+pin