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Popular >

((2cos(x)+1))/(2sin(x)-sqrt(3))>0

Solution

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

Solution

\frac{π}{3} + 2 πn < x < \frac{2 π}{3} + 2 πn or \frac{2 π}{3} + 2 πn < x < \frac{4 π}{3} + 2 πn

Interval Notation

(\frac{π}{3} + 2 πn, \frac{2 π}{3} + 2 πn) \cup (\frac{2 π}{3} + 2 πn, \frac{4 π}{3} + 2 πn)

Decimal Notation

1.04719 \dots + 2 πn < x < 2.09439 \dots + 2 πn or 2.09439 \dots + 2 πn < x < 4.18879 \dots + 2 πn

Solution steps

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

Periodicity of

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

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

is composed of the following functions and periods:

cos (x)

with periodicity of

2 π

The compound periodicity is:

= 2 π

Find the zeroes and undifined points of

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

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

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

\frac{2 cos ( x ) + 1}{2 sin ( x ) - 3} = 0, 0 \leq x < 2 π : x = \frac{4 π}{3}

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

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

2 cos (x) + 1 = 0

Move

1

to the right side

2 cos (x) + 1 = 0

Subtract

1

from both sides

2 cos (x) + 1 - 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{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

x = \frac{2 π}{3} + 2 πn, x = \frac{4 π}{3} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{2 π}{3}, x = \frac{4 π}{3}

Since the equation is undefined for:

\frac{2 π}{3}

x = \frac{4 π}{3}

Find the undefined points:

x = \frac{π}{3}, x = \frac{2 π}{3}

Find the zeros of the denominator

2 sin (x) - 3 = 0

Move

3

to the right side

2 sin (x) - 3 = 0

Add

3

to both sides

2 sin (x) - 3 + 3 = 0 + 3

Simplify

2 sin (x) = 3

2 sin (x) = 3

Divide both sides by

2

2 sin (x) = 3

Divide both sides by

2

\frac{2 sin ( x )}{2} = \frac{3}{2}

Simplify

sin (x) = \frac{3}{2}

sin (x) = \frac{3}{2}

General solutions for

sin (x) = \frac{3}{2}

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

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

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{3}, x = \frac{2 π}{3}

\frac{π}{3}, \frac{2 π}{3}, \frac{4 π}{3}

Identify the intervals

0 < x < \frac{π}{3}, \frac{π}{3} < x < \frac{2 π}{3}, \frac{2 π}{3} < x < \frac{4 π}{3}, \frac{4 π}{3} < x < 2 π

Summarize in a table:

2 cos (x) + 1 2 sin (x) - 3 \frac{2 c o s ( x ) + 1}{2 s i n ( x ) - 3} x = 0 + - - 0 < x < \frac{π}{3} + - - x = \frac{π}{3} + 0 Undefined \frac{π}{3} < x < \frac{2 π}{3} + + + x = \frac{2 π}{3} 00 Undefined \frac{2 π}{3} < x < \frac{4 π}{3} - - + x = \frac{4 π}{3} 0 - 0 \frac{4 π}{3} < x < 2 π + - - x = 2 π + - -

Identify the intervals that satisfy the required condition:

> 0

\frac{π}{3} < x < \frac{2 π}{3} or \frac{2 π}{3} < x < \frac{4 π}{3}

Apply the periodicity of

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

\frac{π}{3} + 2 πn < x < \frac{2 π}{3} + 2 πn or \frac{2 π}{3} + 2 πn < x < \frac{4 π}{3} + 2 πn

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