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Popular >

solvefor x,(10)/(cos(x))>10

Solution

solve for

x, \frac{10}{cos ( x )} > 10

Solution

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

Solution steps

\frac{10}{cos ( x )} > 10

Rewrite in standard form

\frac{10}{cos ( x )} > 10

Subtract

10

from both sides

\frac{10}{cos ( x )} - 10 > 10 - 10

Simplify

\frac{10}{cos ( x )} - 10 > 0

Simplify

\frac{10}{cos ( x )} - 10 : \frac{10 - 10 cos ( x )}{cos ( x )}

\frac{10}{cos ( x )} - 10

Convert element to fraction:

10 = \frac{10 cos ( x )}{cos ( x )}

= \frac{10}{cos ( x )} - \frac{10 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{10 - 10 cos ( x )}{cos ( x )}

\frac{10 - 10 cos ( x )}{cos ( x )} > 0

\frac{10 - 10 cos ( x )}{cos ( x )} > 0

Factor

\frac{10 - 10 cos ( x )}{cos ( x )} : \frac{- 10 ( cos ( x ) - 1 )}{cos ( x )}

\frac{10 - 10 cos ( x )}{cos ( x )}

Factor

- 10 cos (x) + 10 : - 10 (cos (x) - 1)

- 10 cos (x) + 10

Factor out common term

- 10

= - 10 (cos (x) - 1)

= \frac{- 10 ( cos ( x ) - 1 )}{cos ( x )}

\frac{- 10 ( cos ( x ) - 1 )}{cos ( x )} > 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - 10 ( cos ( x ) - 1 ) ) ( - 1 )}{cos ( x )} < 0 \cdot (- 1)

Simplify

\frac{10 ( cos ( x ) - 1 )}{cos ( x )} < 0

Divide both sides by

10

\frac{\frac{10 ( c o s ( x ) - 1 )}{c o s ( x )}}{10} < \frac{0}{10}

Simplify

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

Identify the intervals

Find the signs of the factors of

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

Find the signs of

cos (x) - 1

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

cos (x) - 1 = 0

Move

1

to the right side

cos (x) - 1 = 0

Add

1

to both sides

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

Simplify

cos (x) = 1

cos (x) = 1

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

cos (x) - 1 < 0

Move

1

to the right side

cos (x) - 1 < 0

Add

1

to both sides

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

Simplify

cos (x) < 1

cos (x) < 1

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

cos (x) - 1 > 0

Move

1

to the right side

cos (x) - 1 > 0

Add

1

to both sides

cos (x) - 1 + 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:

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

Identify the intervals that satisfy the required condition:

< 0

0 < cos (x) < 1

0 < cos (x) < 1

a < u < b

then

a < u and u < b

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

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) < 1 : 2 πn < x < 2 π + 2 πn

cos (x) < 1

For

cos (x) < a

, if

- 1 < a \leq 1

then

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

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

Simplify

arccos (1) : 0

arccos (1)

Use the following trivial identity:

arccos (1) = 0

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}

= 0

Simplify

2 π - arccos (1) : 2 π

2 π - arccos (1)

Use the following trivial identity:

arccos (1) = 0

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 - 0

2 π - 0 = 2 π

= 2 π

0 + 2 πn < x < 2 π + 2 πn

Simplify

2 πn < x < 2 π + 2 πn

Combine the intervals

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

Merge Overlapping Intervals

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

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