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Popular Trigonometry >

(2sin(x)-1)(-2cos(x)+sqrt(2))<= 0

Solution

(2 sin (x) - 1) (- 2 cos (x) + 2) \leq 0

Solution

\frac{π}{6} + 2 πn \leq x \leq \frac{π}{4} + 2 πn or \frac{5 π}{6} + 2 πn \leq x \leq \frac{7 π}{4} + 2 πn

Interval Notation

[\frac{π}{6} + 2 πn, \frac{π}{4} + 2 πn] \cup [\frac{5 π}{6} + 2 πn, \frac{7 π}{4} + 2 πn]

Decimal

0.52359 \dots + 2 πn \leq x \leq 0.78539 \dots + 2 πn or 2.61799 \dots + 2 πn \leq x \leq 5.49778 \dots + 2 πn

Solution steps

(2 sin (x) - 1) (- 2 cos (x) + 2) \leq 0

Periodicity of

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

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

is composed of the following functions and periods:

sin (x)

with periodicity of

2 π

The compound periodicity is:

= 2 π

To find the zeroes, set the inequality to zero

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

Solve

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

for

0 \leq x < 2 π

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

Solving each part separately

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}

- 2 cos (x) + 2 = 0 : x = \frac{π}{4} or x = \frac{7 π}{4}

- 2 cos (x) + 2 = 0, 0 \leq x < 2 π

Move

2

to the right side

- 2 cos (x) + 2 = 0

Subtract

2

from both sides

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

Simplify

- 2 cos (x) = - 2

- 2 cos (x) = - 2

Divide both sides by

- 2

- 2 cos (x) = - 2

Divide both sides by

- 2

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

Simplify

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

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

General solutions for

cos (x) = \frac{2}{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}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

Solutions for the range

0 \leq x < 2 π

x = \frac{π}{4}, x = \frac{7 π}{4}

Combine all the solutions

\frac{π}{6} or \frac{π}{4} or \frac{5 π}{6} or \frac{7 π}{4}

The intervals between the zeros

0 < x < \frac{π}{6}, \frac{π}{6} < x < \frac{π}{4}, \frac{π}{4} < x < \frac{5 π}{6}, \frac{5 π}{6} < x < \frac{7 π}{4}, \frac{7 π}{4} < x < 2 π

Summarize in a table:

2 sin (x) - 1 - 2 cos (x) + 2 (2 sin (x) - 1) (- 2 cos (x) + 2) x = 0 - - + 0 < x < \frac{π}{6} - - + x = \frac{π}{6} 0 - 0 \frac{π}{6} < x < \frac{π}{4} + - - x = \frac{π}{4} + 00 \frac{π}{4} < x < \frac{5 π}{6} + + + x = \frac{5 π}{6} 0 + 0 \frac{5 π}{6} < x < \frac{7 π}{4} - + - x = \frac{7 π}{4} - 00 \frac{7 π}{4} < x < 2 π - - + x = 2 π - - +

Identify the intervals that satisfy the required condition:

\leq 0

x = \frac{π}{6} or \frac{π}{6} < x < \frac{π}{4} or x = \frac{π}{4} or x = \frac{5 π}{6} or \frac{5 π}{6} < x < \frac{7 π}{4} or x = \frac{7 π}{4}

Merge Overlapping Intervals

\frac{π}{6} \leq x \leq \frac{π}{4} or \frac{5 π}{6} \leq x < \frac{7 π}{4} or x = \frac{7 π}{4}