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Popular Trigonometry >

cos(2x)<= sin(x)

Solution

cos (2 x) \leq sin (x)

Solution

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

Interval Notation

[\frac{π}{6} + 2 πn, \frac{5 π}{6} + 2 πn] \cup x = - \frac{π}{2} + 2 πn

Decimal

0.52359 \dots + 2 πn \leq x \leq 2.61799 \dots + 2 πn or x = - 1.57079 \dots + 2 πn

Solution steps

cos (2 x) \leq sin (x)

Move

sin (x)

to the left side

cos (2 x) \leq sin (x)

Subtract

sin (x)

from both sides

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

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

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

Use the following identity:

cos (2 x) = 1 - 2 sin^{2} (x)

1 - 2 sin^{2} (x) - sin (x) \leq 0

Let:

u = sin (x)

1 - 2 u^{2} - u \leq 0

1 - 2 u^{2} - u \leq 0 : u \leq - 1 or u \geq \frac{1}{2}

1 - 2 u^{2} - u \leq 0

Factor

1 - 2 u^{2} - u : - (2 u - 1) (u + 1)

1 - 2 u^{2} - u

Factor out common term

- 1

= - (2 u^{2} + u - 1)

Factor

2 u^{2} + u - 1 : (2 u - 1) (u + 1)

2 u^{2} + u - 1

Write in the standard form

a x^{2} + b x + c

= 2 u^{2} + u - 1

Break the expression into groups

2 u^{2} + u - 1

Definition

Factors of

2 : 1, 2

2

Divisors (Factors)

Find the Prime factors of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Add 1

1

The factors of

2

1, 2

Negative factors of

2 : - 1, - 2

Multiply the factors by

- 1

to get the negative factors

- 1, - 2

For every two factors such that

u * v = - 2,

check if

u + v = 1

Check

u = 1, v = - 2 : u * v = - 2, u + v = - 1 \Rightarrow

FalseCheck

u = 2, v = - 1 : u * v = - 2, u + v = 1 \Rightarrow

True

u = 2, v = - 1

Group into

(a x^{2} + ux) + (vx + c)

(2 u^{2} - u) + (2 u - 1)

= (2 u^{2} - u) + (2 u - 1)

Factor out

u

from

2 u^{2} - u : u (2 u - 1)

2 u^{2} - u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{2} = uu

= 2 uu - u

Factor out common term

u

= u (2 u - 1)

= u (2 u - 1) + (2 u - 1)

Factor out common term

2 u - 1

= (2 u - 1) (u + 1)

= - (2 u - 1) (u + 1)

- (2 u - 1) (u + 1) \leq 0

Multiply both sides by

- 1

(reverse the inequality)

(- (2 u - 1) (u + 1)) (- 1) \geq 0 \cdot (- 1)

Simplify

(2 u - 1) (u + 1) \geq 0

Identify the intervals

Find the signs of the factors of

(2 u - 1) (u + 1)

Find the signs of

2 u - 1

2 u - 1 = 0 : u = \frac{1}{2}

2 u - 1 = 0

Move

1

to the right side

2 u - 1 = 0

Add

1

to both sides

2 u - 1 + 1 = 0 + 1

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

2 u - 1 < 0 : u < \frac{1}{2}

2 u - 1 < 0

Move

1

to the right side

2 u - 1 < 0

Add

1

to both sides

2 u - 1 + 1 < 0 + 1

Simplify

2 u < 1

2 u < 1

Divide both sides by

2

2 u < 1

Divide both sides by

2

\frac{2 u}{2} < \frac{1}{2}

Simplify

u < \frac{1}{2}

u < \frac{1}{2}

2 u - 1 > 0 : u > \frac{1}{2}

2 u - 1 > 0

Move

1

to the right side

2 u - 1 > 0

Add

1

to both sides

2 u - 1 + 1 > 0 + 1

Simplify

2 u > 1

2 u > 1

Divide both sides by

2

2 u > 1

Divide both sides by

2

\frac{2 u}{2} > \frac{1}{2}

Simplify

u > \frac{1}{2}

u > \frac{1}{2}

Find the signs of

u + 1

u + 1 = 0 : u = - 1

u + 1 = 0

Move

1

to the right side

u + 1 = 0

Subtract

1

from both sides

u + 1 - 1 = 0 - 1

Simplify

u = - 1

u = - 1

u + 1 < 0 : u < - 1

u + 1 < 0

Move

1

to the right side

u + 1 < 0

Subtract

1

from both sides

u + 1 - 1 < 0 - 1

Simplify

u < - 1

u < - 1

u + 1 > 0 : u > - 1

u + 1 > 0

Move

1

to the right side

u + 1 > 0

Subtract

1

from both sides

u + 1 - 1 > 0 - 1

Simplify

u > - 1

u > - 1

Summarize in a table:

2 u - 1 u + 1 (2 u - 1) (u + 1) u < - 1 - - + u = - 1 - 00 - 1 < u < \frac{1}{2} - + - u = \frac{1}{2} 0 + 0 u > \frac{1}{2} + + +

Identify the intervals that satisfy the required condition:

\geq 0

u < - 1 or u = - 1 or u = \frac{1}{2} or u > \frac{1}{2}

Merge Overlapping Intervals

u \leq - 1 or u = \frac{1}{2} or u > \frac{1}{2}

The union of two intervals is the set of numbers which are in either interval

u < - 1

u = - 1

u \leq - 1

The union of two intervals is the set of numbers which are in either interval

u \leq - 1

u = \frac{1}{2}

u \leq - 1 or u = \frac{1}{2}

The union of two intervals is the set of numbers which are in either interval

u \leq - 1 or u = \frac{1}{2}

u > \frac{1}{2}

u \leq - 1 or u \geq \frac{1}{2}

u \leq - 1 or u \geq \frac{1}{2}

u \leq - 1 or u \geq \frac{1}{2}

u \leq - 1 or u \geq \frac{1}{2}

Substitute back

u = sin (x)

sin (x) \leq - 1 or sin (x) \geq \frac{1}{2}

sin (x) \leq - 1 : x = - \frac{π}{2} + 2 πn

sin (x) \leq - 1

For

sin (x) \leq a

, if

- 1 < a < 1

then

- π - arcsin (a) + 2 πn \leq x \leq arcsin (a) + 2 πn

- π - arcsin (- 1) + 2 πn \leq x \leq arcsin (- 1) + 2 πn

Simplify

- π - arcsin (- 1) : - \frac{π}{2}

- π - arcsin (- 1)

arcsin (- 1) = - \frac{π}{2}

arcsin (- 1)

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- 1) = - arcsin (1)

= - arcsin (1)

Use the following trivial identity:

arcsin (1) = \frac{π}{2}

arcsin (1)

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{2}

= - \frac{π}{2}

= - π - (- \frac{π}{2})

Simplify

- π - (- \frac{π}{2})

Apply rule

- (- a) = a

= - π + \frac{π}{2}

Convert element to fraction:

π = \frac{π 2}{2}

= - \frac{π 2}{2} + \frac{π}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 2 + π}{2}

Add similar elements:

- 2 π + π = - π

= \frac{- π}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{π}{2}

= - \frac{π}{2}

Simplify

arcsin (- 1) : - \frac{π}{2}

arcsin (- 1)

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- 1) = - arcsin (1)

= - arcsin (1)

Use the following trivial identity:

arcsin (1) = \frac{π}{2}

arcsin (1)

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{2}

= - \frac{π}{2}

- \frac{π}{2} + 2 πn \leq x \leq - \frac{π}{2} + 2 πn

Simplify

x = - \frac{π}{2} + 2 πn

sin (x) \geq \frac{1}{2} : \frac{π}{6} + 2 πn \leq x \leq \frac{5 π}{6} + 2 πn

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

For

sin (x) \geq a

, if

- 1 < a < 1

then

arcsin (a) + 2 πn \leq x \leq π - arcsin (a) + 2 πn

arcsin (\frac{1}{2}) + 2 πn \leq x \leq π - arcsin (\frac{1}{2}) + 2 πn

Simplify

arcsin (\frac{1}{2}) : \frac{π}{6}

arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{6}

Simplify

π - arcsin (\frac{1}{2}) : \frac{5 π}{6}

π - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{6}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= π - \frac{π}{6}

Simplify

π - \frac{π}{6}

Convert element to fraction:

π = \frac{π 6}{6}

= \frac{π 6}{6} - \frac{π}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{π 6 - π}{6}

Add similar elements:

6 π - π = 5 π

= \frac{5 π}{6}

= \frac{5 π}{6}

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

Combine the intervals

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

Merge Overlapping Intervals

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

Popular Examples

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cos(x)-1/2 cos(2x)>0

sec(x)<= sqrt(2)

1/(cos(x))<= 2,-pi<= x<= pi

sin(x)>0.5