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

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

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2csc(2x)cot(2x)=0

2 csc (2 x) cot (2 x) = 0

x = \frac{π}{4} + \frac{πn}{2}

Degrees

x = 4 5^{\circ} + 9 0^{\circ} n

Solution steps

2 csc (2 x) cot (2 x) = 0

Solving each part separately

csc (2 x) = 0 or cot (2 x) = 0

csc (2 x) = 0 :

No Solution

csc (2 x) = 0

csc (x) \leq - 1 or csc (x) \geq 1

No Solution

cot (2 x) = 0 : x = \frac{π}{4} + \frac{πn}{2}

cot (2 x) = 0

General solutions for

cot (2 x) = 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 x = \frac{π}{2} + πn

2 x = \frac{π}{2} + πn

Solve

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

2 x = \frac{π}{2} + πn

Divide both sides by

2

2 x = \frac{π}{2} + πn

Divide both sides by

2

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

Simplify

\frac{2 x}{2} = \frac{\frac{π}{2}}{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{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}

x = \frac{π}{4} + \frac{πn}{2}

x = \frac{π}{4} + \frac{πn}{2}

x = \frac{π}{4} + \frac{πn}{2}

x = \frac{π}{4} + \frac{πn}{2}

Combine all the solutions

x = \frac{π}{4} + \frac{πn}{2}

3 cos (y) + cot (y) = 0

2 sin (x) - 3 sin (x) + 1 = 0

2 sin (x) (sin (x) - 1) = 3 cos^{2} (x)

2 sin^{2} (x) = 2 sin (x) + 4

sin (x) cos^{2} (x) = sin (x), 0 \leq x \leq 2 π

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