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Popular >

sec(x)<-1

Solution

sec (x) < - 1

Solution

\frac{π}{2} + 2 πn < x < π + 2 πn or π + 2 πn < x < \frac{3 π}{2} + 2 πn

Interval Notation

(\frac{π}{2} + 2 πn, π + 2 πn) \cup (π + 2 πn, \frac{3 π}{2} + 2 πn)

Decimal Notation

1.57079 \dots + 2 πn < x < 3.14159 \dots + 2 πn or 3.14159 \dots + 2 πn < x < 4.71238 \dots + 2 πn

Solution steps

sec (x) < - 1

Express with sin, cos

sec (x) < - 1

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

\frac{1}{cos ( x )} < - 1

\frac{1}{cos ( x )} < - 1

Rewrite in standard form

\frac{1}{cos ( x )} < - 1

Add

1

to both sides

\frac{1}{cos ( x )} + 1 < - 1 + 1

Simplify

\frac{1}{cos ( x )} + 1 < 0

Simplify

\frac{1}{cos ( x )} + 1 : \frac{1 + cos ( x )}{cos ( x )}

\frac{1}{cos ( x )} + 1

Convert element to fraction:

1 = \frac{1 cos ( x )}{cos ( x )}

= \frac{1}{cos ( x )} + \frac{1 \cdot cos ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 + 1 \cdot cos ( x )}{cos ( x )}

Multiply:

1 \cdot cos (x) = cos (x)

= \frac{1 + cos ( x )}{cos ( x )}

\frac{1 + cos ( x )}{cos ( x )} < 0

\frac{1 + cos ( x )}{cos ( x )} < 0

Identify the intervals

Find the signs of the factors of

\frac{1 + cos ( x )}{cos ( x )}

Find the signs of

1 + cos (x)

1 + cos (x) = 0 : cos (x) = - 1

1 + cos (x) = 0

Move

1

to the right side

1 + cos (x) = 0

Subtract

1

from both sides

1 + cos (x) - 1 = 0 - 1

Simplify

cos (x) = - 1

cos (x) = - 1

1 + cos (x) < 0 : cos (x) < - 1

1 + cos (x) < 0

Move

1

to the right side

1 + cos (x) < 0

Subtract

1

from both sides

1 + cos (x) - 1 < 0 - 1

Simplify

cos (x) < - 1

cos (x) < - 1

1 + cos (x) > 0 : cos (x) > - 1

1 + cos (x) > 0

Move

1

to the right side

1 + cos (x) > 0

Subtract

1

from both sides

1 + cos (x) - 1 > 0 - 1

Simplify

cos (x) > - 1

cos (x) > - 1

Find the signs of

cos (x)

cos (x) = 0

cos (x) < 0

cos (x) > 0

Find singularity points

Find the zeros of the denominator

cos (x) : cos (x) = 0

Summarize in a table:

1 + cos (x) cos (x) \frac{1 + c o s ( x )}{c o s ( x )} cos (x) < - 1 - - + cos (x) = - 1 0 - 0 - 1 < cos (x) < 0 + - - cos (x) = 0 + 0 Undefined cos (x) > 0 + + +

Identify the intervals that satisfy the required condition:

< 0

- 1 < cos (x) < 0

- 1 < cos (x) < 0

a < u < b

then

a < u and u < b

- 1 < cos (x) and cos (x) < 0

- 1 < cos (x) : - π + 2 πn < x < π + 2 πn

- 1 < cos (x)

Switch sides

cos (x) > - 1

For

cos (x) > a

, if

- 1 \leq a < 1

then

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (- 1) + 2 πn < x < arccos (- 1) + 2 πn

Simplify

- arccos (- 1) : - π

- arccos (- 1)

Use the following trivial identity:

arccos (- 1) = π

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - π

Simplify

arccos (- 1) : π

arccos (- 1)

Use the following trivial identity:

arccos (- 1) = π

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= π

- π + 2 πn < x < π + 2 πn

cos (x) < 0 : \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

cos (x) < 0

For

cos (x) < a

, if

- 1 < a \leq 1

then

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

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

Simplify

arccos (0) : \frac{π}{2}

arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2}

Simplify

2 π - arccos (0) : \frac{3 π}{2}

2 π - arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{2}

Simplify

2 π - \frac{π}{2}

Convert element to fraction:

2 π = \frac{2 π 2}{2}

= \frac{2 π 2}{2} - \frac{π}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 2 - π}{2}

2 π 2 - π = 3 π

2 π 2 - π

Multiply the numbers:

2 \cdot 2 = 4

= 4 π - π

Add similar elements:

4 π - π = 3 π

= 3 π

= \frac{3 π}{2}

= \frac{3 π}{2}

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

Combine the intervals

- π + 2 πn < x < π + 2 πn and \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

Merge Overlapping Intervals

\frac{π}{2} + 2 πn < x < π + 2 πn or π + 2 πn < x < \frac{3 π}{2} + 2 πn

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