What is the general solution for 4cot(2θ)=0 ?

The general solution for 4cot(2θ)=0 is θ= pi/4+(pin)/2

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4cot(2θ)=0

4 cot (2 θ) = 0

θ = \frac{π}{4} + \frac{πn}{2}

Degrees

θ = 4 5^{\circ} + 9 0^{\circ} n

Solution steps

4 cot (2 θ) = 0

Divide both sides by

4

4 cot (2 θ) = 0

Divide both sides by

4

\frac{4 cot ( 2 θ )}{4} = \frac{0}{4}

Simplify

cot (2 θ) = 0

cot (2 θ) = 0

General solutions for

cot (2 θ) = 0

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}{2} + πn

2 θ = \frac{π}{2} + πn

Solve

2 θ = \frac{π}{2} + πn : θ = \frac{π}{4} + \frac{πn}{2}

2 θ = \frac{π}{2} + πn

Divide both sides by

2

2 θ = \frac{π}{2} + πn

Divide both sides by

2

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

Simplify

\frac{2 θ}{2} = \frac{\frac{π}{2}}{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}{2}}{2} + \frac{πn}{2} : \frac{π}{4} + \frac{πn}{2}

\frac{\frac{π}{2}}{2} + \frac{πn}{2}

\frac{\frac{π}{2}}{2} = \frac{π}{4}

\frac{\frac{π}{2}}{2}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{π}{4}

= \frac{π}{4} + \frac{πn}{2}

θ = \frac{π}{4} + \frac{πn}{2}

θ = \frac{π}{4} + \frac{πn}{2}

θ = \frac{π}{4} + \frac{πn}{2}

θ = \frac{π}{4} + \frac{πn}{2}

sin (6 0^{\circ}) = cos (x + 1 0^{\circ})

sin (2 x) = 682

0 = 5 \cdot sin (2 π \cdot (- 0.25 - x))

cos (a) = \frac{2}{2}

csc (x) = 1.1223

What is the general solution for 4cot(2θ)=0 ?
The general solution for 4cot(2θ)=0 is θ= pi/4+(pin)/2