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

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

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cot(2θ)=(sqrt(3))/3

cot (2 θ) = \frac{3}{3}

θ = \frac{π}{6} + \frac{πn}{2}

Degrees

θ = 3 0^{\circ} + 9 0^{\circ} n

Solution steps

cot (2 θ) = \frac{3}{3}

General solutions for

cot (2 θ) = \frac{3}{3}

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 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}

4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0, 0 \leq θ \leq 36 0^{\circ}

- cos (θ) + 1 = 2 cos (θ) + 3

cot^{2} (x) + cot (x) = 0

3 cos (2 θ) + 9 cos (θ) - 4 = - 2 cos (θ)

16 cos^{2} (x) - 12 = 0

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