What is the general solution for sqrt(3)tan(2θ)+1=0 ?

The general solution for sqrt(3)tan(2θ)+1=0 is θ=(5pi}{12}+\frac{pin)/2

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Popular >

sqrt(3)tan(2θ)+1=0

Solution

3 tan (2 θ) + 1 = 0

Solution

θ = \frac{5 π}{12} + \frac{πn}{2}

Degrees

θ = 7 5^{\circ} + 9 0^{\circ} n

Solution steps

3 tan (2 θ) + 1 = 0

Move

1

to the right side

3 tan (2 θ) + 1 = 0

Subtract

1

from both sides

3 tan (2 θ) + 1 - 1 = 0 - 1

Simplify

3 tan (2 θ) = - 1

3 tan (2 θ) = - 1

Divide both sides by

3

3 tan (2 θ) = - 1

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 tan ( 2 θ )}{3} : tan (2 θ)

\frac{3 tan ( 2 θ )}{3}

Cancel the common factor:

3

= tan (2 θ)

Simplify

\frac{- 1}{3} : - \frac{3}{3}

\frac{- 1}{3}

Apply the fraction rule:

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

= - \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 (2 θ) = - \frac{3}{3}

tan (2 θ) = - \frac{3}{3}

tan (2 θ) = - \frac{3}{3}

General solutions for

tan (2 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}

2 θ = \frac{5 π}{6} + πn

2 θ = \frac{5 π}{6} + πn

Solve

2 θ = \frac{5 π}{6} + πn : θ = \frac{5 π}{12} + \frac{πn}{2}

2 θ = \frac{5 π}{6} + πn

Divide both sides by

2

2 θ = \frac{5 π}{6} + πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{5 π}{6}}{2} + \frac{πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{5 π}{6}}{2} + \frac{πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{5 π}{6}}{2} + \frac{πn}{2} : \frac{5 π}{12} + \frac{πn}{2}

\frac{\frac{5 π}{6}}{2} + \frac{πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

= \frac{5 π}{12} + \frac{πn}{2}

θ = \frac{5 π}{12} + \frac{πn}{2}

θ = \frac{5 π}{12} + \frac{πn}{2}

θ = \frac{5 π}{12} + \frac{πn}{2}

θ = \frac{5 π}{12} + \frac{πn}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)tan(2θ)+1=0 ?
The general solution for sqrt(3)tan(2θ)+1=0 is θ=(5pi}{12}+\frac{pin)/2