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Popular >

sin(x)> 1/(sin(x))

Solution

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

Solution

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

Interval Notation

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

Decimal Notation

3.14159 \dots + 2 πn < x < 4.71238 \dots + 2 πn or 4.71238 \dots + 2 πn < x < 6.28318 \dots + 2 πn

Solution steps

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

Let:

u = sin (x)

u > \frac{1}{u}

u > \frac{1}{u} : - 1 < u < 0 or u > 1

u > \frac{1}{u}

Rewrite in standard form

u > \frac{1}{u}

Subtract

\frac{1}{u}

from both sides

u - \frac{1}{u} > \frac{1}{u} - \frac{1}{u}

Simplify

u - \frac{1}{u} > 0

Simplify

u - \frac{1}{u} : \frac{u ^{2} - 1}{u}

u - \frac{1}{u}

Convert element to fraction:

u = \frac{uu}{u}

= \frac{uu}{u} - \frac{1}{u}

Since the denominators are equal, combine the fractions:

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

= \frac{uu - 1}{u}

uu - 1 = u^{2} - 1

uu - 1

uu = u^{2}

uu

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

uu = u^{1 + 1}

= u^{1 + 1}

Add the numbers:

1 + 1 = 2

= u^{2}

= u^{2} - 1

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

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

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

Factor

\frac{u ^{2} - 1}{u} : \frac{( u + 1 ) ( u - 1 )}{u}

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

Factor

u^{2} - 1 : (u + 1) (u - 1)

u^{2} - 1

Rewrite

1

1^{2}

= u^{2} - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

u^{2} - 1^{2} = (u + 1) (u - 1)

= (u + 1) (u - 1)

= \frac{( u + 1 ) ( u - 1 )}{u}

\frac{( u + 1 ) ( u - 1 )}{u} > 0

Identify the intervals

Find the signs of the factors of

\frac{( u + 1 ) ( u - 1 )}{u}

Find the signs of

u + 1

u + 1 = 0 : u = - 1

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

u + 1 < 0 : u < - 1

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

u + 1 > 0 : u > - 1

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

Find the signs of

u - 1

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to both sides

u - 1 + 1 = 0 + 1

Simplify

u = 1

u = 1

u - 1 < 0 : u < 1

u - 1 < 0

Move

1

to the right side

u - 1 < 0

Add

1

to both sides

u - 1 + 1 < 0 + 1

Simplify

u < 1

u < 1

u - 1 > 0 : u > 1

u - 1 > 0

Move

1

to the right side

u - 1 > 0

Add

1

to both sides

u - 1 + 1 > 0 + 1

Simplify

u > 1

u > 1

Find the signs of

u

u = 0

u < 0

u > 0

Find singularity points

Find the zeros of the denominator

u : u = 0

Summarize in a table:

u + 1 u - 1 u \frac{( u + 1 ) ( u - 1 )}{u} u < - 1 - - - - u = - 1 0 - - 0 - 1 < u < 0 + - - + u = 0 + - 0 Undefined 0 < u < 1 + - + - u = 1 + 0 + 0 u > 1 + + + +

Identify the intervals that satisfy the required condition:

> 0

- 1 < u < 0 or u > 1

- 1 < u < 0 or u > 1

- 1 < u < 0 or u > 1

Substitute back

u = sin (x)

- 1 < sin (x) < 0 or sin (x) > 1

- 1 < sin (x) < 0 : π + 2 πn < x < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn

- 1 < sin (x) < 0

a < u < b

then

a < u and u < b

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

- 1 < sin (x) : - \frac{π}{2} + 2 πn < x < \frac{3 π}{2} + 2 πn

- 1 < sin (x)

Switch sides

sin (x) > - 1

For

sin (x) > a

, if

- 1 \leq a < 1

then

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

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

Simplify

arcsin (- 1) : - \frac{π}{2}

arcsin (- 1)

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- 1) = - arcsin (1)

= - arcsin (1)

Use the following trivial identity:

arcsin (1) = \frac{π}{2}

arcsin (1)

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}{2}

= - \frac{π}{2}

Simplify

π - arcsin (- 1) : \frac{3 π}{2}

π - arcsin (- 1)

arcsin (- 1) = - \frac{π}{2}

arcsin (- 1)

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- 1) = - arcsin (1)

= - arcsin (1)

Use the following trivial identity:

arcsin (1) = \frac{π}{2}

arcsin (1)

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}{2}

= - \frac{π}{2}

= π - (- \frac{π}{2})

Simplify

π - (- \frac{π}{2})

Apply rule

- (- a) = a

= π + \frac{π}{2}

Convert element to fraction:

π = \frac{π 2}{2}

= \frac{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{π 2 + π}{2}

Add similar elements:

2 π + π = 3 π

= \frac{3 π}{2}

= \frac{3 π}{2}

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

Merge Overlapping Intervals

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

sin (x) > 1 :

False for all

x \in R

sin (x) > 1

Range of

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

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (x) \leq 1

- 1 \leq sin (x) \leq 1

sin (x) > 1 and - 1 \leq sin (x) \leq 1 :

False

Let

y = sin (x)

Combine the intervals

y > 1 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y > 1 and - 1 \leq y \leq 1

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

y > 1

and

- 1 \leq y \leq 1

False for all y \in R

False for all y \in R

No Solution for x \in R

False for all x \in R

Combine the intervals

(π + 2 πn < x < \frac{3 π}{2} + 2 πn or \frac{3 π}{2} + 2 πn < x < 2 π + 2 πn) or False for all x \in R

Merge Overlapping Intervals

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

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