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

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

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tan(2θ)=sqrt(3)

tan (2 θ) = 3

θ = \frac{π}{6} + \frac{πn}{2}

Degrees

θ = 3 0^{\circ} + 9 0^{\circ} n

Solution steps

tan (2 θ) = 3

General solutions for

tan (2 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}

2 θ = \frac{π}{3} + πn

2 θ = \frac{π}{3} + πn

Solve

2 θ = \frac{π}{3} + πn : θ = \frac{π}{6} + \frac{πn}{2}

2 θ = \frac{π}{3} + πn

Divide both sides by

2

2 θ = \frac{π}{3} + πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{π}{3}}{2} + \frac{πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{π}{3}}{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}{3}}{2} + \frac{πn}{2} : \frac{π}{6} + \frac{πn}{2}

\frac{\frac{π}{3}}{2} + \frac{πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{π}{6}

= \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

cos (x) = 0.2

sin (2 x) - \frac{3}{2} = 0

csc^{2} (θ) - 4 = 0

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

cos (θ) = 0.5

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