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

The general solution for 3tan(2θ)-sqrt(3)=0 is θ= pi/(12)+(pin)/2

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3tan(2θ)-sqrt(3)=0

3 tan (2 θ) - 3 = 0

θ = \frac{π}{12} + \frac{πn}{2}

Degrees

θ = 1 5^{\circ} + 9 0^{\circ} n

Solution steps

3 tan (2 θ) - 3 = 0

Move

3

to the right side

3 tan (2 θ) - 3 = 0

Add

3

to both sides

3 tan (2 θ) - 3 + 3 = 0 + 3

Simplify

3 tan (2 θ) = 3

3 tan (2 θ) = 3

Divide both sides by

3

3 tan (2 θ) = 3

Divide both sides by

3

\frac{3 tan ( 2 θ )}{3} = \frac{3}{3}

Simplify

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{π}{6} + πn

2 θ = \frac{π}{6} + πn

Solve

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

2 θ = \frac{π}{6} + πn

Divide both sides by

2

2 θ = \frac{π}{6} + πn

Divide both sides by

2

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

Simplify

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

\frac{\frac{π}{6}}{2} + \frac{πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

= \frac{π}{12} + \frac{πn}{2}

θ = \frac{π}{12} + \frac{πn}{2}

θ = \frac{π}{12} + \frac{πn}{2}

θ = \frac{π}{12} + \frac{πn}{2}

θ = \frac{π}{12} + \frac{πn}{2}

cos (θ) = \frac{4 k - 1}{3}

5 sin (x) - 4 = 0

tan^{2} (θ) - 2 tan (θ) - 3 = 0

cos^{3} (x) + sin^{2} (x) \cdot cos (x) = - \frac{1}{2}

cos (3 A) + sin (A) = 0

What is the general solution for 3tan(2θ)-sqrt(3)=0 ?
The general solution for 3tan(2θ)-sqrt(3)=0 is θ= pi/(12)+(pin)/2