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sin(2x)>= cos(x)

Solution

sin (2 x) \geq cos (x)

Solution

\frac{π}{6} + 2 πn \leq x \leq \frac{π}{2} + 2 πn or \frac{5 π}{6} + 2 πn \leq x \leq \frac{3 π}{2} + 2 πn

Interval Notation

[\frac{π}{6} + 2 πn, \frac{π}{2} + 2 πn] \cup [\frac{5 π}{6} + 2 πn, \frac{3 π}{2} + 2 πn]

Decimal Notation

0.52359 \dots + 2 πn \leq x \leq 1.57079 \dots + 2 πn or 2.61799 \dots + 2 πn \leq x \leq 4.71238 \dots + 2 πn

Solution steps

sin (2 x) \geq cos (x)

Move

cos (x)

to the left side

sin (2 x) \geq cos (x)

Subtract

cos (x)

from both sides

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

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

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

Use the following identity:

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

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

Periodicity of

- cos (x) + 2 cos (x) sin (x) : 2 π

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

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

Periodicity of

cos (x) : 2 π

Periodicity of

cos (x)

2 π

= 2 π

Periodicity of

2 cos (x) sin (x) : π

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

Combine periods:

2 π, π

= 2 π

Factor

- cos (x) + 2 cos (x) sin (x) : cos (x) (2 sin (x) - 1)

- cos (x) + 2 cos (x) sin (x)

Factor out common term

cos (x)

= cos (x) (- 1 + 2 sin (x))

cos (x) (2 sin (x) - 1) \geq 0

To find the zeroes, set the inequality to zero

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

Solve

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

for

0 \leq x < 2 π

cos (x) (2 sin (x) - 1) = 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) - 1 = 0 : x = \frac{π}{6} or x = \frac{5 π}{6}

2 sin (x) - 1 = 0, 0 \leq x < 2 π

Move

1

to the right side

2 sin (x) - 1 = 0

Add

1

to both sides

2 sin (x) - 1 + 1 = 0 + 1

Simplify

2 sin (x) = 1

2 sin (x) = 1

Divide both sides by

2

2 sin (x) = 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

sin (x) = \frac{1}{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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{6}, x = \frac{5 π}{6}

Combine all the solutions

\frac{π}{6} or \frac{π}{2} or \frac{5 π}{6} or \frac{3 π}{2}

The intervals between the zeros

0 < x < \frac{π}{6}, \frac{π}{6} < x < \frac{π}{2}, \frac{π}{2} < x < \frac{5 π}{6}, \frac{5 π}{6} < x < \frac{3 π}{2}, \frac{3 π}{2} < x < 2 π

Summarize in a table:

cos (x) 2 sin (x) - 1 cos (x) (2 sin (x) - 1) x = 0 + - - 0 < x < \frac{π}{6} + - - x = \frac{π}{6} + 00 \frac{π}{6} < x < \frac{π}{2} + + + x = \frac{π}{2} 0 + 0 \frac{π}{2} < x < \frac{5 π}{6} - + - x = \frac{5 π}{6} - 00 \frac{5 π}{6} < 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}{6} or \frac{π}{6} < x < \frac{π}{2} or x = \frac{π}{2} or x = \frac{5 π}{6} or \frac{5 π}{6} < x < \frac{3 π}{2} or x = \frac{3 π}{2}

Merge Overlapping Intervals

\frac{π}{6} \leq x \leq \frac{π}{2} or \frac{5 π}{6} \leq x < \frac{3 π}{2} or x = \frac{3 π}{2}