What is the general solution for 2cot(2x)+2=0 ?

The general solution for 2cot(2x)+2=0 is x=(3pi)/8+(pin)/2

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2cot(2x)+2=0

2 cot (2 x) + 2 = 0

x = \frac{3 π}{8} + \frac{πn}{2}

Degrees

x = 67. 5^{\circ} + 9 0^{\circ} n

Solution steps

2 cot (2 x) + 2 = 0

Move

2

to the right side

2 cot (2 x) + 2 = 0

Subtract

2

from both sides

2 cot (2 x) + 2 - 2 = 0 - 2

Simplify

2 cot (2 x) = - 2

2 cot (2 x) = - 2

Divide both sides by

2

2 cot (2 x) = - 2

Divide both sides by

2

\frac{2 cot ( 2 x )}{2} = \frac{- 2}{2}

Simplify

cot (2 x) = - 1

cot (2 x) = - 1

General solutions for

cot (2 x) = - 1

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 x = \frac{3 π}{4} + πn

2 x = \frac{3 π}{4} + πn

Solve

2 x = \frac{3 π}{4} + πn : x = \frac{3 π}{8} + \frac{πn}{2}

2 x = \frac{3 π}{4} + πn

Divide both sides by

2

2 x = \frac{3 π}{4} + πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{3 π}{4}}{2} + \frac{πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{3 π}{4}}{2} + \frac{πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{3 π}{4}}{2} + \frac{πn}{2} : \frac{3 π}{8} + \frac{πn}{2}

\frac{\frac{3 π}{4}}{2} + \frac{πn}{2}

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

Apply the fraction rule:

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

= \frac{3 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{3 π}{8}

= \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

sin (x - π) = \frac{- 2}{2}

sec^{2} (θ) tan (θ) = 2 tan (θ), 0^{\circ} \leq θ \leq 36 0^{\circ}

arccos (x) = 18 0^{\circ}

sin (a) = \frac{29.7}{54.5}

1.0003 sin (3 0^{\circ}) = 1.46 sin (x)

What is the general solution for 2cot(2x)+2=0 ?
The general solution for 2cot(2x)+2=0 is x=(3pi)/8+(pin)/2