What is the general solution for -sqrt(3)tan(θ)-6=-3 ?

The general solution for -sqrt(3)tan(θ)-6=-3 is θ=(2pi)/3+pin

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-sqrt(3)tan(θ)-6=-3

- 3 tan (θ) - 6 = - 3

θ = \frac{2 π}{3} + πn

Degrees

θ = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

- 3 tan (θ) - 6 = - 3

Move

6

to the right side

- 3 tan (θ) - 6 = - 3

Add

6

to both sides

- 3 tan (θ) - 6 + 6 = - 3 + 6

Simplify

- 3 tan (θ) = 3

- 3 tan (θ) = 3

Divide both sides by

- 3

- 3 tan (θ) = 3

Divide both sides by

- 3

\frac{- 3 tan ( θ )}{- 3} = \frac{3}{- 3}

Simplify

\frac{- 3 tan ( θ )}{- 3} = \frac{3}{- 3}

Simplify

\frac{- 3 tan ( θ )}{- 3} : tan (θ)

\frac{- 3 tan ( θ )}{- 3}

Apply the fraction rule:

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

= \frac{3 tan ( θ )}{3}

Cancel the common factor:

3

= tan (θ)

Simplify

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

\frac{3}{- 3}

Apply the fraction rule:

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

= - \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

tan (θ) = - 3

tan (θ) = - 3

tan (θ) = - 3

General solutions for

tan (θ) = - 3

tan (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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{2 π}{3} + πn

θ = \frac{2 π}{3} + πn

tan (θ) = 2.5

arccos (1 - x) + arccos (x) = arccos (- x)

tan (x) = 0.75

2 sin (x) tan (x) = tan (x)

4 csc (θ) + 5 = 0

What is the general solution for -sqrt(3)tan(θ)-6=-3 ?
The general solution for -sqrt(3)tan(θ)-6=-3 is θ=(2pi)/3+pin