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Popular >

1/(sin(2x))< 1/(sin(x))

Solution

\frac{1}{sin ( 2 x )} < \frac{1}{sin ( x )}

Solution

2 πn < x < \frac{π}{3} + 2 πn or \frac{π}{2} + 2 πn < x < π + 2 πn or \frac{3 π}{2} + 2 πn < x < \frac{5 π}{3} + 2 πn

Interval Notation

(2 πn, \frac{π}{3} + 2 πn) \cup (\frac{π}{2} + 2 πn, π + 2 πn) \cup (\frac{3 π}{2} + 2 πn, \frac{5 π}{3} + 2 πn)

Decimal Notation

2 πn < x < 1.04719 \dots + 2 πn or 1.57079 \dots + 2 πn < x < 3.14159 \dots + 2 πn or 4.71238 \dots + 2 πn < x < 5.23598 \dots + 2 πn

Solution steps

\frac{1}{sin ( 2 x )} < \frac{1}{sin ( x )}

Move

\frac{1}{sin ( x )}

to the left side

\frac{1}{sin ( 2 x )} < \frac{1}{sin ( x )}

Subtract

\frac{1}{sin ( x )}

from both sides

\frac{1}{sin ( 2 x )} - \frac{1}{sin ( x )} < \frac{1}{sin ( x )} - \frac{1}{sin ( x )}

\frac{1}{sin ( 2 x )} - \frac{1}{sin ( x )} < 0

\frac{1}{sin ( 2 x )} - \frac{1}{sin ( x )} < 0

Use the following identity:

sin (2 x) = 2 cos (x) sin (x)

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

Simplify

\frac{1}{2 cos ( x ) sin ( x )} - \frac{1}{sin ( x )} : \frac{1 - 2 cos ( x )}{2 cos ( x ) sin ( x )}

\frac{1}{2 cos ( x ) sin ( x )} - \frac{1}{sin ( x )}

Least Common Multiplier of

2 cos (x) sin (x), sin (x) : 2 cos (x) sin (x)

2 cos (x) sin (x), sin (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

2 cos (x) sin (x)

sin (x)

= 2 cos (x) sin (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

2 cos (x) sin (x)

For

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

multiply the denominator and numerator by

2 cos (x)

\frac{1}{sin ( x )} = \frac{1 \cdot 2 cos ( x )}{sin ( x ) \cdot 2 cos ( x )} = \frac{2 cos ( x )}{2 cos ( x ) sin ( x )}

= \frac{1}{2 cos ( x ) sin ( x )} - \frac{2 cos ( x )}{2 cos ( x ) sin ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{1 - 2 cos ( x )}{2 cos ( x ) sin ( x )}

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

Periodicity of

\frac{1}{2 cos ( x ) sin ( x )} - \frac{1}{sin ( x )} : 2 π

The compound periodicity of the sum of periodic functions is the least common multiplier of the periods

\frac{1}{2 cos ( x ) sin ( x )}, \frac{1}{sin ( x )}

Periodicity of

\frac{1}{2 cos ( x ) sin ( x )} : π

\frac{1}{2 cos ( x ) sin ( x )}

is composed of the following functions and periods:

cos (x)

with periodicity of

2 π

The compound periodicity is:

π

Periodicity of

\frac{1}{sin ( x )} : 2 π

Periodicity of

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

Periodicity of

sin (x)

2 π

= \frac{2 π}{∣ 1 ∣}

Simplify

= 2 π

Combine periods:

π, 2 π

= 2 π

Find the zeroes and undifined points of

\frac{1 - 2 cos ( x )}{2 cos ( x ) sin ( x )}

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

\frac{1 - 2 cos ( x )}{2 cos ( x ) sin ( x )} = 0, 0 \leq x < 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

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

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

1 - 2 cos (x) = 0

Move

1

to the right side

1 - 2 cos (x) = 0

Subtract

1

from both sides

1 - 2 cos (x) - 1 = 0 - 1

Simplify

- 2 cos (x) = - 1

- 2 cos (x) = - 1

Divide both sides by

- 2

- 2 cos (x) = - 1

Divide both sides by

- 2

\frac{- 2 cos ( x )}{- 2} = \frac{- 1}{- 2}

Simplify

cos (x) = \frac{1}{2}

cos (x) = \frac{1}{2}

General solutions for

cos (x) = \frac{1}{2}

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}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

Find the undefined points:

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

Find the zeros of the denominator

2 cos (x) sin (x) = 0

Solving each part separately

cos (x) = 0 or sin (x) = 0

cos (x) = 0, 0 \leq x < 2 π : x = \frac{π}{2}, x = \frac{3 π}{2}

cos (x) = 0, 0 \leq x < 2 π

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}

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

General solutions for

sin (x) = 0

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

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Solutions for the range

0 \leq x < 2 π

x = 0, x = π

Combine all the solutions

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

0, \frac{π}{3}, \frac{π}{2}, π, \frac{3 π}{2}, \frac{5 π}{3}

Identify the intervals

0 < x < \frac{π}{3}, \frac{π}{3} < x < \frac{π}{2}, \frac{π}{2} < x < π, π < x < \frac{3 π}{2}, \frac{3 π}{2} < x < \frac{5 π}{3}, \frac{5 π}{3} < x < 2 π

Summarize in a table:

1 - 2 cos (x) cos (x) sin (x) \frac{1 - 2 c o s ( x )}{2 c o s ( x ) s i n ( x )} x = 0 - + 0 Undefined 0 < x < \frac{π}{3} - + + - x = \frac{π}{3} 0 + + 0 \frac{π}{3} < x < \frac{π}{2} + + + + x = \frac{π}{2} + 0 + Undefined \frac{π}{2} < x < π + - + - x = π + - 0 Undefined π < x < \frac{3 π}{2} + - - + x = \frac{3 π}{2} + 0 - Undefined \frac{3 π}{2} < x < \frac{5 π}{3} + + - - x = \frac{5 π}{3} 0 + - 0 \frac{5 π}{3} < x < 2 π - + - + x = 2 π - + 0 Undefined

Identify the intervals that satisfy the required condition:

< 0

0 < x < \frac{π}{3} or \frac{π}{2} < x < π or \frac{3 π}{2} < x < \frac{5 π}{3}

Apply the periodicity of

\frac{1}{2 cos ( x ) sin ( x )} - \frac{1}{sin ( x )}

2 πn < x < \frac{π}{3} + 2 πn or \frac{π}{2} + 2 πn < x < π + 2 πn or \frac{3 π}{2} + 2 πn < x < \frac{5 π}{3} + 2 πn

Popular Examples

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