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Popular Trigonometry >

-2+3csc(2x-pi)>= 0

Solution

- 2 + 3 csc (2 x - π) \geq 0

Solution

\frac{π}{2} + πn < x < π + πn

Interval Notation

(\frac{π}{2} + πn, π + πn)

Decimal

1.57079 \dots + πn < x < 3.14159 \dots + πn

Solution steps

- 2 + 3 csc (2 x - π) \geq 0

Periodicity of

- 2 + 3 csc (2 x - π) : π

Periodicity of

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

Periodicity of

csc (x)

2 π

= \frac{2 π}{∣ 2 ∣}

Simplify

= π

Express with sin, cos

- 2 + 3 csc (2 x - π) \geq 0

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

- 2 + 3 \cdot \frac{1}{sin ( 2 x - π )} \geq 0

- 2 + 3 \cdot \frac{1}{sin ( 2 x - π )} \geq 0

Simplify

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

- 2 + 3 \cdot \frac{1}{sin ( 2 x - π )}

3 \cdot \frac{1}{sin ( 2 x - π )} = \frac{3}{sin ( 2 x - π )}

3 \cdot \frac{1}{sin ( 2 x - π )}

Multiply fractions:

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

= \frac{1 \cdot 3}{sin ( 2 x - π )}

Multiply the numbers:

1 \cdot 3 = 3

= \frac{3}{sin ( 2 x - π )}

= - 2 + \frac{3}{sin ( 2 x - π )}

Convert element to fraction:

2 = \frac{2 sin ( 2 x - π )}{sin ( 2 x - π )}

= - \frac{2 sin ( 2 x - π )}{sin ( 2 x - π )} + \frac{3}{sin ( 2 x - π )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 2 sin ( 2 x - π ) + 3}{sin ( 2 x - π )}

\frac{- 2 sin ( 2 x - π ) + 3}{sin ( 2 x - π )} \geq 0

Find the zeroes and undifined points of

\frac{- 2 sin ( 2 x - π ) + 3}{sin ( 2 x - π )}

for

0 \leq x < π

To find the zeroes, set the inequality to zero

\frac{- 2 sin ( 2 x - π ) + 3}{sin ( 2 x - π )} = 0

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

No Solution for

x \in R

\frac{- 2 sin ( 2 x - π ) + 3}{sin ( 2 x - π )} = 0, 0 \leq x < π

Solve by substitution

\frac{- 2 sin ( 2 x - π ) + 3}{sin ( 2 x - π )} = 0

Let:

sin (2 x - π) = u

\frac{- 2 u + 3}{u} = 0

\frac{- 2 u + 3}{u} = 0 : u = \frac{3}{2}

\frac{- 2 u + 3}{u} = 0

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

- 2 u + 3 = 0

Move

3

to the right side

- 2 u + 3 = 0

Subtract

3

from both sides

- 2 u + 3 - 3 = 0 - 3

Simplify

- 2 u = - 3

- 2 u = - 3

Divide both sides by

- 2

- 2 u = - 3

Divide both sides by

- 2

\frac{- 2 u}{- 2} = \frac{- 3}{- 2}

Simplify

u = \frac{3}{2}

u = \frac{3}{2}

Verify Solutions

Find undefined (singularity) points:

u = 0

Take the denominator(s) of

\frac{- 2 u + 3}{u}

and compare to zero

u = 0

The following points are undefined

u = 0

Combine undefined points with solutions:

u = \frac{3}{2}

Substitute back

u = sin (2 x - π)

sin (2 x - π) = \frac{3}{2}

sin (2 x - π) = \frac{3}{2}

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

No Solution

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

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution for x \in R

Find the undefined points:

x = \frac{π}{2}, x = 0

Find the zeros of the denominator

sin (2 x - π) = 0

General solutions for

sin (2 x - π) = 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}

2 x - π = 0 + 2 πn, 2 x - π = π + 2 πn

2 x - π = 0 + 2 πn, 2 x - π = π + 2 πn

Solve

2 x - π = 0 + 2 πn : x = πn + \frac{π}{2}

2 x - π = 0 + 2 πn

0 + 2 πn = 2 πn

2 x - π = 2 πn

Move

π

to the right side

2 x - π = 2 πn

Add

π

to both sides

2 x - π + π = 2 πn + π

Simplify

2 x = 2 πn + π

2 x = 2 πn + π

Divide both sides by

2

2 x = 2 πn + π

Divide both sides by

2

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

Simplify

x = πn + \frac{π}{2}

x = πn + \frac{π}{2}

Solve

2 x - π = π + 2 πn : x = π + πn

2 x - π = π + 2 πn

Move

π

to the right side

2 x - π = π + 2 πn

Add

π

to both sides

2 x - π + π = π + 2 πn + π

Simplify

2 x = 2 π + 2 πn

2 x = 2 π + 2 πn

Divide both sides by

2

2 x = 2 π + 2 πn

Divide both sides by

2

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

Simplify

x = π + πn

x = π + πn

x = πn + \frac{π}{2}, x = π + πn

Solutions for the range

0 \leq x < π

x = \frac{π}{2}, x = 0

0, \frac{π}{2}

Identify the intervals

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

Summarize in a table:

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

Identify the intervals that satisfy the required condition:

\geq 0

\frac{π}{2} < x < π

Apply the periodicity of

- 2 + 3 csc (2 x - π)

\frac{π}{2} + πn < x < π + πn

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