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Popular Trigonometry >

1-1/(cos(x))< 1/2*10^{-2}

Solution

1 - \frac{1}{cos ( x )} < \frac{1}{2} \cdot 1 0^{- 2}

Solution

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

Interval Notation

(- \frac{π}{2} + 2 πn, \frac{π}{2} + 2 πn)

Decimal

- 1.57079 \dots + 2 πn < x < 1.57079 \dots + 2 πn

Solution steps

1 - \frac{1}{cos ( x )} < \frac{1}{2} \cdot 1 0^{- 2}

Rewrite in standard form

1 - \frac{1}{cos ( x )} < \frac{1}{2} \cdot 1 0^{- 2}

Subtract

\frac{1}{2} 1 0^{- 2}

from both sides

1 - \frac{1}{cos ( x )} - \frac{1}{2} \cdot 1 0^{- 2} < \frac{1}{2} \cdot 1 0^{- 2} - \frac{1}{2} \cdot 1 0^{- 2}

Simplify

1 - \frac{1}{cos ( x )} - \frac{1}{2} \cdot 1 0^{- 2} < \frac{1}{2} \cdot 1 0^{- 2} - \frac{1}{2} \cdot 1 0^{- 2}

Simplify

1 - \frac{1}{cos ( x )} - \frac{1}{2} \cdot 1 0^{- 2} : 1 - \frac{1}{cos ( x )} - \frac{1}{200}

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

\frac{1}{2} \cdot 1 0^{- 2} = \frac{1}{200}

\frac{1}{2} \cdot 1 0^{- 2}

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

1 0^{- 2} = \frac{1}{1 0 ^{2}}

= \frac{1}{2} \cdot \frac{1}{1 0 ^{2}}

Multiply fractions:

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

= \frac{1 \cdot 1}{2 \cdot 1 0 ^{2}}

Multiply the numbers:

1 \cdot 1 = 1

= \frac{1}{1 0 ^{2} \cdot 2}

2 \cdot 1 0^{2} = 200

2 \cdot 1 0^{2}

1 0^{2} = 100

= 2 \cdot 100

Multiply the numbers:

2 \cdot 100 = 200

= 200

= \frac{1}{200}

= 1 - \frac{1}{cos ( x )} - \frac{1}{200}

1 - \frac{1}{cos ( x )} - \frac{1}{200} < 0

Simplify

1 - \frac{1}{cos ( x )} - \frac{1}{200} : \frac{199 cos ( x ) - 200}{200 cos ( x )}

1 - \frac{1}{cos ( x )} - \frac{1}{200}

Convert element to fraction:

1 = \frac{1}{1}

= \frac{1}{1} - \frac{1}{cos ( x )} - \frac{1}{200}

Least Common Multiplier of

1, cos (x), 200 : 200 cos (x)

1, cos (x), 200

Lowest Common Multiplier (LCM)

Least Common Multiplier of

1, 200 : 200

1, 200

Least Common Multiplier (LCM)

Prime factorization of

1

Prime factorization of

200 : 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5

200

200

divides by

2200 = 100 \cdot 2

= 2 \cdot 100

100

divides by

2100 = 50 \cdot 2

= 2 \cdot 2 \cdot 50

50

divides by

250 = 25 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 25

25

divides by

525 = 5 \cdot 5

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5

2, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5

Multiply each factor the greatest number of times it occurs in either

1

200

= 2 \cdot 2 \cdot 2 \cdot 5 \cdot 5

Multiply the numbers:

2 \cdot 2 \cdot 2 \cdot 5 \cdot 5 = 200

= 200

Compute an expression comprised of factors that appear in at least one of the factored expressions

= 200 cos (x)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

200 cos (x)

For

\frac{1}{1} :

multiply the denominator and numerator by

200 cos (x)

\frac{1}{1} = \frac{1 \cdot 200 cos ( x )}{1 \cdot 200 cos ( x )} = \frac{200 cos ( x )}{200 cos ( x )}

For

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

multiply the denominator and numerator by

200

\frac{1}{cos ( x )} = \frac{1 \cdot 200}{cos ( x ) \cdot 200} = \frac{200}{200 cos ( x )}

For

\frac{1}{200} :

multiply the denominator and numerator by

cos (x)

\frac{1}{200} = \frac{1 \cdot cos ( x )}{200 cos ( x )} = \frac{cos ( x )}{200 cos ( x )}

= \frac{200 cos ( x )}{200 cos ( x )} - \frac{200}{200 cos ( x )} - \frac{cos ( x )}{200 cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{200 cos ( x ) - 200 - cos ( x )}{200 cos ( x )}

200 cos (x) - 200 - cos (x) = 199 cos (x) - 200

200 cos (x) - 200 - cos (x)

Group like terms

= 200 cos (x) - cos (x) - 200

Add similar elements:

200 cos (x) - cos (x) = 199 cos (x)

= 199 cos (x) - 200

= \frac{199 cos ( x ) - 200}{200 cos ( x )}

\frac{199 cos ( x ) - 200}{200 cos ( x )} < 0

Multiply both sides by

200

\frac{200 ( 199 cos ( x ) - 200 )}{200 cos ( x )} < 0 \cdot 200

Simplify

\frac{199 cos ( x ) - 200}{cos ( x )} < 0

\frac{199 cos ( x ) - 200}{cos ( x )} < 0

Identify the intervals

Find the signs of the factors of

\frac{199 cos ( x ) - 200}{cos ( x )}

Find the signs of

199 cos (x) - 200

199 cos (x) - 200 = 0 : cos (x) = \frac{200}{199}

199 cos (x) - 200 = 0

Move

200

to the right side

199 cos (x) - 200 = 0

Add

200

to both sides

199 cos (x) - 200 + 200 = 0 + 200

Simplify

199 cos (x) = 200

199 cos (x) = 200

Divide both sides by

199

199 cos (x) = 200

Divide both sides by

199

\frac{199 cos ( x )}{199} = \frac{200}{199}

Simplify

cos (x) = \frac{200}{199}

cos (x) = \frac{200}{199}

199 cos (x) - 200 < 0 : cos (x) < \frac{200}{199}

199 cos (x) - 200 < 0

Move

200

to the right side

199 cos (x) - 200 < 0

Add

200

to both sides

199 cos (x) - 200 + 200 < 0 + 200

Simplify

199 cos (x) < 200

199 cos (x) < 200

Divide both sides by

199

199 cos (x) < 200

Divide both sides by

199

\frac{199 cos ( x )}{199} < \frac{200}{199}

Simplify

cos (x) < \frac{200}{199}

cos (x) < \frac{200}{199}

199 cos (x) - 200 > 0 : cos (x) > \frac{200}{199}

199 cos (x) - 200 > 0

Move

200

to the right side

199 cos (x) - 200 > 0

Add

200

to both sides

199 cos (x) - 200 + 200 > 0 + 200

Simplify

199 cos (x) > 200

199 cos (x) > 200

Divide both sides by

199

199 cos (x) > 200

Divide both sides by

199

\frac{199 cos ( x )}{199} > \frac{200}{199}

Simplify

cos (x) > \frac{200}{199}

cos (x) > \frac{200}{199}

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:

199 cos (x) - 200 cos (x) \frac{199 c o s ( x ) - 200}{c o s ( x )} cos (x) < 0 - - + cos (x) = 0 - 0 Undefined 0 < cos (x) < \frac{200}{199} - + - cos (x) = \frac{200}{199} 0 + 0 cos (x) > \frac{200}{199} + + +

Identify the intervals that satisfy the required condition:

< 0

0 < cos (x) < \frac{200}{199}

0 < cos (x) < \frac{200}{199}

a < u < b

then

a < u and u < b

0 < cos (x) and cos (x) < \frac{200}{199}

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

0 < cos (x)

Switch sides

cos (x) > 0

For

cos (x) > a

, if

- 1 \leq a < 1

then

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

- arccos (0) + 2 πn < x < 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

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}

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

cos (x) < \frac{200}{199} :

True for all

x \in R

cos (x) < \frac{200}{199}

Range of

cos (x) : - 1 \leq cos (x) \leq 1

Function range definition

The range of the basic

cos

function is

- 1 \leq cos (x) \leq 1

- 1 \leq cos (x) \leq 1

cos (x) < \frac{200}{199} and - 1 \leq cos (x) \leq 1 : - 1 \leq cos (x) \leq 1

Let

y = cos (x)

Combine the intervals

y < \frac{200}{199} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y < \frac{200}{199} and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y < \frac{200}{199}

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all x

True for all x \in R

Combine the intervals

- \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn and True for all x \in R

Merge Overlapping Intervals

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

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