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cos(x)(2sin(x)-sqrt(3))>= 0

Solution

cos (x) (2 sin (x) - 3) \geq 0

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

Interval Notation

[\frac{π}{3} + 2 πn, \frac{π}{2} + 2 πn] \cup [\frac{2 π}{3} + 2 πn, \frac{3 π}{2} + 2 πn]

Decimal Notation

1.04719 \dots + 2 πn \leq x \leq 1.57079 \dots + 2 πn or 2.09439 \dots + 2 πn \leq x \leq 4.71238 \dots + 2 πn

Solution steps

cos (x) (2 sin (x) - 3) \geq 0

Periodicity of

cos (x) (2 sin (x) - 3) : 2 π

cos (x) (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

To find the zeroes, set the inequality to zero

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

Solve

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

for

0 \leq x < 2 π

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

Solving each part separately

cos (x) = 0 : x = \frac{π}{2} or 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}

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

2 sin (x) - 3 = 0, 0 \leq x < 2 π

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}

Combine all the solutions

\frac{π}{3} or \frac{π}{2} or \frac{2 π}{3} or \frac{3 π}{2}

The intervals between the zeros

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

Summarize in a table:

cos (x) 2 sin (x) - 3 cos (x) (2 sin (x) - 3) x = 0 + - - 0 < x < \frac{π}{3} + - - x = \frac{π}{3} + 00 \frac{π}{3} < x < \frac{π}{2} + + + x = \frac{π}{2} 0 + 0 \frac{π}{2} < x < \frac{2 π}{3} - + - x = \frac{2 π}{3} - 00 \frac{2 π}{3} < x < \frac{3 π}{2} - - + x = \frac{3 π}{2} 0 - 0 \frac{3 π}{2} < x < 2 π + - - x = 2 π + - -

Identify the intervals that satisfy the required condition:

\geq 0

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

Merge Overlapping Intervals

\frac{π}{3} \leq x \leq \frac{π}{2} or \frac{2 π}{3} \leq x < \frac{3 π}{2} or x = \frac{3 π}{2}