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Popular >

1+sec(x)>= 0

Solution

1 + sec (x) \geq 0

Solution

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

Interval Notation

[2 πn, \frac{π}{2} + 2 πn) \cup x = π + 2 πn \cup (\frac{3 π}{2} + 2 πn, 2 π + 2 πn]

Decimal Notation

2 πn \leq x < 1.57079 \dots + 2 πn or x = 3.14159 \dots + 2 πn or 4.71238 \dots + 2 πn < x \leq 6.28318 \dots + 2 πn

Solution steps

1 + sec (x) \geq 0

Periodicity of

1 + sec (x) : 2 π

Periodicity of

a \cdot sec (b x + c) + d = \frac{periodicity of sec ( x )}{∣ b ∣}

Periodicity of

sec (x)

2 π

= \frac{2 π}{∣ 1 ∣}

Simplify

= 2 π

Express with sin, cos

1 + sec (x) \geq 0

Use the basic trigonometric identity:

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

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

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

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

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

Find the zeroes and undifined points of

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

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

\frac{cos ( x ) + 1}{cos ( x )} = 0, 0 \leq x < 2 π : x = π

\frac{cos ( x ) + 1}{cos ( x )} = 0, 0 \leq x < 2 π

Solve by substitution

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

Let:

cos (x) = u

\frac{u + 1}{u} = 0

\frac{u + 1}{u} = 0 : u = - 1

\frac{u + 1}{u} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

\frac{u + 1}{u}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = - 1

Substitute back

u = cos (x)

cos (x) = - 1

cos (x) = - 1

cos (x) = - 1, 0 \leq x < 2 π : x = π

cos (x) = - 1, 0 \leq x < 2 π

General solutions for

cos (x) = - 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

x = π + 2 πn

x = π + 2 πn

Solutions for the range

0 \leq x < 2 π

x = π

Combine all the solutions

x = π

Find the undefined points:

x = \frac{π}{2}, x = \frac{3 π}{2}

Find the zeros of the denominator

cos (x) = 0

General solutions for

cos (x) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

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

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

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{2}, x = \frac{3 π}{2}

\frac{π}{2}, π, \frac{3 π}{2}

Identify the intervals

0 < x < \frac{π}{2}, \frac{π}{2} < x < π, π < x < \frac{3 π}{2}, \frac{3 π}{2} < x < 2 π

Summarize in a table:

cos (x) + 1 cos (x) \frac{c o s ( x ) + 1}{c o s ( x )} x = 0 + + + 0 < x < \frac{π}{2} + + + x = \frac{π}{2} + 0 Undefined \frac{π}{2} < x < π + - - x = π 0 - 0 π < x < \frac{3 π}{2} + - - x = \frac{3 π}{2} + 0 Undefined \frac{3 π}{2} < x < 2 π + + + x = 2 π + + +

Identify the intervals that satisfy the required condition:

\geq 0

x = 0 or 0 < x < \frac{π}{2} or x = π or \frac{3 π}{2} < x < 2 π or x = 2 π

Merge Overlapping Intervals

0 \leq x < \frac{π}{2} or x = π or \frac{3 π}{2} < x < 2 π or x = 2 π

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

x = 0

0 < x < \frac{π}{2}

0 \leq x < \frac{π}{2}

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

0 \leq x < \frac{π}{2}

x = π

0 \leq x < \frac{π}{2} or x = π

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

0 \leq x < \frac{π}{2} or x = π

\frac{3 π}{2} < x < 2 π

0 \leq x < \frac{π}{2} or x = π or \frac{3 π}{2} < x < 2 π

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

0 \leq x < \frac{π}{2} or x = π or \frac{3 π}{2} < x < 2 π

x = 2 π

0 \leq x < \frac{π}{2} or x = π or \frac{3 π}{2} < x \leq 2 π

0 \leq x < \frac{π}{2} or x = π or \frac{3 π}{2} < x \leq 2 π

Apply the periodicity of

1 + sec (x)

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

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