What is the general solution for 3-1/2*tan(θ)=(6+sqrt(3))/2 ?

The general solution for 3-1/2*tan(θ)=(6+sqrt(3))/2 is θ=(2pi)/3+pin

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3-1/2*tan(θ)=(6+sqrt(3))/2

Solution

3 - \frac{1}{2} \cdot tan (θ) = \frac{6 + 3}{2}

Solution

θ = \frac{2 π}{3} + πn

Degrees

θ = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

3 - \frac{1}{2} tan (θ) = \frac{6 + 3}{2}

Move

3

to the right side

3 - \frac{1}{2} tan (θ) = \frac{6 + 3}{2}

Subtract

3

from both sides

3 - \frac{1}{2} tan (θ) - 3 = \frac{6 + 3}{2} - 3

Simplify

- \frac{1}{2} tan (θ) = \frac{6 + 3}{2} - 3

- \frac{1}{2} tan (θ) = \frac{6 + 3}{2} - 3

Multiply both sides by

- 2

- \frac{1}{2} tan (θ) = \frac{6 + 3}{2} - 3

Multiply both sides by

- 2

(- \frac{1}{2} tan (θ)) (- 2) = \frac{6 + 3}{2} (- 2) - 3 (- 2)

Simplify

(- \frac{1}{2} tan (θ)) (- 2) = \frac{6 + 3}{2} (- 2) - 3 (- 2)

Simplify

(- \frac{1}{2} tan (θ)) (- 2) : tan (θ)

(- \frac{1}{2} tan (θ)) (- 2)

Remove parentheses:

(- a) = - a, - (- a) = a

= \frac{1}{2} tan (θ) \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2} tan (θ)

Cancel the common factor:

2

= tan (θ) \cdot 1

Multiply:

tan (θ) \cdot 1 = tan (θ)

= tan (θ)

Simplify

\frac{6 + 3}{2} (- 2) - 3 (- 2) : - 3

\frac{6 + 3}{2} (- 2) - 3 (- 2)

Remove parentheses:

(- a) = - a, - (- a) = a

= - \frac{6 + 3}{2} \cdot 2 + 3 \cdot 2

\frac{6 + 3}{2} \cdot 2 = 6 + 3

\frac{6 + 3}{2} \cdot 2

Multiply fractions:

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

= \frac{( 6 + 3 ) \cdot 2}{2}

Cancel the common factor:

2

= 6 + 3

3 \cdot 2 = 6

3 \cdot 2

Multiply the numbers:

3 \cdot 2 = 6

= 6

= - (6 + 3) + 6

- (6 + 3) : - 6 - 3

- (6 + 3)

Distribute parentheses

= - (6) - (3)

Apply minus-plus rules

+ (- a) = - a

= - 6 - 3

= - 6 - 3 + 6

- 6 + 6 = 0

= - 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

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Frequently Asked Questions (FAQ)

What is the general solution for 3-1/2*tan(θ)=(6+sqrt(3))/2 ?
The general solution for 3-1/2*tan(θ)=(6+sqrt(3))/2 is θ=(2pi)/3+pin