What is the general solution for -7sqrt(3)tan(θ)+2=9 ?

The general solution for -7sqrt(3)tan(θ)+2=9 is θ=(5pi)/6+pin

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Popular >

-7sqrt(3)tan(θ)+2=9

Solution

- 73 tan (θ) + 2 = 9

Solution

θ = \frac{5 π}{6} + πn

Degrees

θ = 15 0^{\circ} + 18 0^{\circ} n

Solution steps

- 73 tan (θ) + 2 = 9

Move

2

to the right side

- 73 tan (θ) + 2 = 9

Subtract

2

from both sides

- 73 tan (θ) + 2 - 2 = 9 - 2

Simplify

- 73 tan (θ) = 7

- 73 tan (θ) = 7

Divide both sides by

- 73

- 73 tan (θ) = 7

Divide both sides by

- 73

\frac{- 7 3 tan ( θ )}{- 7 3} = \frac{7}{- 7 3}

Simplify

\frac{- 7 3 tan ( θ )}{- 7 3} = \frac{7}{- 7 3}

Simplify

\frac{- 7 3 tan ( θ )}{- 7 3} : tan (θ)

\frac{- 7 3 tan ( θ )}{- 7 3}

Apply the fraction rule:

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

= \frac{7 3 tan ( θ )}{7 3}

Divide the numbers:

\frac{7}{7} = 1

= \frac{3 tan ( θ )}{3}

Cancel the common factor:

3

= tan (θ)

Simplify

\frac{7}{- 7 3} : - \frac{3}{3}

\frac{7}{- 7 3}

Apply the fraction rule:

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

= - \frac{7}{7 3}

Divide the numbers:

\frac{7}{7} = 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}

tan (θ) = - \frac{3}{3}

tan (θ) = - \frac{3}{3}

tan (θ) = - \frac{3}{3}

General solutions for

tan (θ) = - \frac{3}{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{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for -7sqrt(3)tan(θ)+2=9 ?
The general solution for -7sqrt(3)tan(θ)+2=9 is θ=(5pi)/6+pin