What is the general solution for csc(2x)+1=0 ?

The general solution for csc(2x)+1=0 is x=(3pi)/4+pin

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csc(2x)+1=0

csc (2 x) + 1 = 0

x = \frac{3 π}{4} + πn

Degrees

x = 13 5^{\circ} + 18 0^{\circ} n

Solution steps

csc (2 x) + 1 = 0

Move

1

to the right side

csc (2 x) + 1 = 0

Subtract

1

from both sides

csc (2 x) + 1 - 1 = 0 - 1

Simplify

csc (2 x) = - 1

csc (2 x) = - 1

General solutions for

csc (2 x) = - 1

csc (x)

periodicity table with

2 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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

2 x = \frac{3 π}{2} + 2 πn

2 x = \frac{3 π}{2} + 2 πn

Solve

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

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

2

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

Simplify

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

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

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

\frac{\frac{3 π}{2}}{2}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3 π}{4}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

x = \frac{3 π}{4} + πn

cos (5 x) - cos (x) = sin (3 x)

1 = cos (\frac{( π ( t - 22 ) )}{183})

cot (x) = - csc^{2} (x)

3 cos^{2} (θ) + 4 cos (θ) - 4 = 0

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

What is the general solution for csc(2x)+1=0 ?
The general solution for csc(2x)+1=0 is x=(3pi)/4+pin