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Popular >

csc(x)<=-2

Solution

csc (x) \leq - 2

Solution

π + 2 πn < x \leq \frac{7 π}{6} + 2 πn or \frac{11 π}{6} + 2 πn \leq x < 2 π + 2 πn

Interval Notation

(π + 2 πn, \frac{7 π}{6} + 2 πn] \cup [\frac{11 π}{6} + 2 πn, 2 π + 2 πn)

Decimal Notation

3.14159 \dots + 2 πn < x \leq 3.66519 \dots + 2 πn or 5.75958 \dots + 2 πn \leq x < 6.28318 \dots + 2 πn

Solution steps

csc (x) \leq - 2

Express with sin, cos

csc (x) \leq - 2

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

\frac{1}{sin ( x )} \leq - 2

\frac{1}{sin ( x )} \leq - 2

Rewrite in standard form

\frac{1}{sin ( x )} \leq - 2

Add

2

to both sides

\frac{1}{sin ( x )} + 2 \leq - 2 + 2

Simplify

\frac{1}{sin ( x )} + 2 \leq 0

Simplify

\frac{1}{sin ( x )} + 2 : \frac{1 + 2 sin ( x )}{sin ( x )}

\frac{1}{sin ( x )} + 2

Convert element to fraction:

2 = \frac{2 sin ( x )}{sin ( x )}

= \frac{1}{sin ( x )} + \frac{2 sin ( x )}{sin ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 + 2 sin ( x )}{sin ( x )}

\frac{1 + 2 sin ( x )}{sin ( x )} \leq 0

\frac{1 + 2 sin ( x )}{sin ( x )} \leq 0

Identify the intervals

Find the signs of the factors of

\frac{1 + 2 sin ( x )}{sin ( x )}

Find the signs of

1 + 2 sin (x)

1 + 2 sin (x) = 0 : sin (x) = - \frac{1}{2}

1 + 2 sin (x) = 0

Move

1

to the right side

1 + 2 sin (x) = 0

Subtract

1

from both sides

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

Simplify

2 sin (x) = - 1

2 sin (x) = - 1

Divide both sides by

2

2 sin (x) = - 1

Divide both sides by

2

\frac{2 sin ( x )}{2} = \frac{- 1}{2}

Simplify

sin (x) = - \frac{1}{2}

sin (x) = - \frac{1}{2}

1 + 2 sin (x) < 0 : sin (x) < - \frac{1}{2}

1 + 2 sin (x) < 0

Move

1

to the right side

1 + 2 sin (x) < 0

Subtract

1

from both sides

1 + 2 sin (x) - 1 < 0 - 1

Simplify

2 sin (x) < - 1

2 sin (x) < - 1

Divide both sides by

2

2 sin (x) < - 1

Divide both sides by

2

\frac{2 sin ( x )}{2} < \frac{- 1}{2}

Simplify

sin (x) < - \frac{1}{2}

sin (x) < - \frac{1}{2}

1 + 2 sin (x) > 0 : sin (x) > - \frac{1}{2}

1 + 2 sin (x) > 0

Move

1

to the right side

1 + 2 sin (x) > 0

Subtract

1

from both sides

1 + 2 sin (x) - 1 > 0 - 1

Simplify

2 sin (x) > - 1

2 sin (x) > - 1

Divide both sides by

2

2 sin (x) > - 1

Divide both sides by

2

\frac{2 sin ( x )}{2} > \frac{- 1}{2}

Simplify

sin (x) > - \frac{1}{2}

sin (x) > - \frac{1}{2}

Find the signs of

sin (x)

sin (x) = 0

sin (x) < 0

sin (x) > 0

Find singularity points

Find the zeros of the denominator

sin (x) : sin (x) = 0

Summarize in a table:

1 + 2 sin (x) sin (x) \frac{1 + 2 s i n ( x )}{s i n ( x )} sin (x) < - \frac{1}{2} - - + sin (x) = - \frac{1}{2} 0 - 0 - \frac{1}{2} < sin (x) < 0 + - - sin (x) = 0 + 0 Undefined sin (x) > 0 + + +

Identify the intervals that satisfy the required condition:

\leq 0

sin (x) = - \frac{1}{2} or - \frac{1}{2} < sin (x) < 0

Merge Overlapping Intervals

sin (x) = - \frac{1}{2} or - \frac{1}{2} < sin (x) < 0

The union of two intervals is the set of numbers which are in either interval

sin (x) = - \frac{1}{2}

- \frac{1}{2} < sin (x) < 0

- \frac{1}{2} \leq sin (x) < 0

- \frac{1}{2} \leq sin (x) < 0

- \frac{1}{2} \leq sin (x) < 0

a \leq u < b

then

a \leq u and u < b

- \frac{1}{2} \leq sin (x) and sin (x) < 0

- \frac{1}{2} \leq sin (x) : - \frac{π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn

- \frac{1}{2} \leq sin (x)

Switch sides

sin (x) \geq - \frac{1}{2}

For

sin (x) \geq a

, if

- 1 < a < 1

then

arcsin (a) + 2 πn \leq x \leq π - arcsin (a) + 2 πn

arcsin (- \frac{1}{2}) + 2 πn \leq x \leq π - arcsin (- \frac{1}{2}) + 2 πn

Simplify

arcsin (- \frac{1}{2}) : - \frac{π}{6}

arcsin (- \frac{1}{2})

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

= - \frac{π}{6}

Simplify

π - arcsin (- \frac{1}{2}) : \frac{7 π}{6}

π - arcsin (- \frac{1}{2})

arcsin (- \frac{1}{2}) = - \frac{π}{6}

arcsin (- \frac{1}{2})

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

= - \frac{π}{6}

= π - (- \frac{π}{6})

Simplify

π - (- \frac{π}{6})

Apply rule

- (- a) = a

= π + \frac{π}{6}

Convert element to fraction:

π = \frac{π 6}{6}

= \frac{π 6}{6} + \frac{π}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{π 6 + π}{6}

Add similar elements:

6 π + π = 7 π

= \frac{7 π}{6}

= \frac{7 π}{6}

- \frac{π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn

sin (x) < 0 : - π + 2 πn < x < 2 πn

sin (x) < 0

For

sin (x) < a

, if

- 1 < a \leq 1

then

- π - arcsin (a) + 2 πn < x < arcsin (a) + 2 πn

- π - arcsin (0) + 2 πn < x < arcsin (0) + 2 πn

Simplify

- π - arcsin (0) : - π

- π - arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= - π - 0

- π - 0 = - π

= - π

Simplify

arcsin (0) : 0

arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= 0

- π + 2 πn < x < 0 + 2 πn

Simplify

- π + 2 πn < x < 2 πn

Combine the intervals

- \frac{π}{6} + 2 πn \leq x \leq \frac{7 π}{6} + 2 πn and - π + 2 πn < x < 2 πn

Merge Overlapping Intervals

π + 2 πn < x \leq \frac{7 π}{6} + 2 πn or \frac{11 π}{6} + 2 πn \leq x < 2 π + 2 πn

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