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Popular Trigonometry >

cos(x)<= sin^2(x)<= (sqrt(3))/2 sin(x)

Solution

cos (x) \leq sin^{2} (x) \leq \frac{3}{2} sin (x)

Solution

arccos (\frac{5 - 1}{2}) + 2 πn \leq x \leq \frac{π}{3} + 2 πn or \frac{2 π}{3} + 2 πn \leq x \leq π + 2 πn

Interval Notation

[arccos (\frac{5 - 1}{2}) + 2 πn, \frac{π}{3} + 2 πn] \cup [\frac{2 π}{3} + 2 πn, π + 2 πn]

Decimal

0.90455 \dots + 2 πn \leq x \leq 1.04719 \dots + 2 πn or 2.09439 \dots + 2 πn \leq x \leq 3.14159 \dots + 2 πn

Solution steps

cos (x) \leq sin^{2} (x) \leq \frac{3}{2} sin (x)

a \leq u \leq b

then

a \leq u and u \leq b

cos (x) \leq sin^{2} (x) and sin^{2} (x) \leq \frac{3}{2} sin (x)

cos (x) \leq sin^{2} (x) : arccos (\frac{5 - 1}{2}) + 2 πn \leq x \leq 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

cos (x) \leq sin^{2} (x)

Move

sin^{2} (x)

to the left side

cos (x) \leq sin^{2} (x)

Subtract

sin^{2} (x)

from both sides

cos (x) - sin^{2} (x) \leq sin^{2} (x) - sin^{2} (x)

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

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

Use the following identity:

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

Therefore

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

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

Simplify

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

Let:

u = cos (x)

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

u - 1 + u^{2} \leq 0 : \frac{- 5 - 1}{2} \leq u \leq \frac{5 - 1}{2}

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

Complete the square

u - 1 + u^{2} : (u + \frac{1}{2})^{2} - \frac{5}{4}

u - 1 + u^{2}

Write in the standard form

a x^{2} + b x + c

u^{2} + u - 1

Write

u^{2} + u - 1

in the form:

x^{2} + 2 a x + a^{2}

2 a = 1 : a = \frac{1}{2}

2 a = 1

Divide both sides by

2

2 a = 1

Divide both sides by

2

\frac{2 a}{2} = \frac{1}{2}

Simplify

a = \frac{1}{2}

a = \frac{1}{2}

Add and subtract

(\frac{1}{2})^{2}

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

x^{2} + 2 a x + a^{2} = (x + a)^{2}

u^{2} + 1 u + (\frac{1}{2})^{2} = (u + \frac{1}{2})^{2}

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

Simplify

(u + \frac{1}{2})^{2} - \frac{5}{4}

(u + \frac{1}{2})^{2} - \frac{5}{4} \leq 0

Move

\frac{5}{4}

to the right side

(u + \frac{1}{2})^{2} - \frac{5}{4} \leq 0

Add

\frac{5}{4}

to both sides

(u + \frac{1}{2})^{2} - \frac{5}{4} + \frac{5}{4} \leq 0 + \frac{5}{4}

Simplify

(u + \frac{1}{2})^{2} \leq \frac{5}{4}

(u + \frac{1}{2})^{2} \leq \frac{5}{4}

For

u^{n} \leq a

, if

n

is even then

- \frac{5}{4} \leq u + \frac{1}{2} \leq \frac{5}{4}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{5}{4} \leq u + \frac{1}{2} and u + \frac{1}{2} \leq \frac{5}{4}

- \frac{5}{4} \leq u + \frac{1}{2} : u \geq \frac{- 5 - 1}{2}

- \frac{5}{4} \leq u + \frac{1}{2}

Switch sides

u + \frac{1}{2} \geq - \frac{5}{4}

Simplify

\frac{5}{4} : \frac{5}{2}

\frac{5}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{5}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{5}{2}

u + \frac{1}{2} \geq - \frac{5}{2}

Move

\frac{1}{2}

to the right side

u + \frac{1}{2} \geq - \frac{5}{2}

Subtract

\frac{1}{2}

from both sides

u + \frac{1}{2} - \frac{1}{2} \geq - \frac{5}{2} - \frac{1}{2}

Simplify

u + \frac{1}{2} - \frac{1}{2} \geq - \frac{5}{2} - \frac{1}{2}

Simplify

u + \frac{1}{2} - \frac{1}{2} : u

u + \frac{1}{2} - \frac{1}{2}

Add similar elements:

\frac{1}{2} - \frac{1}{2} \geq 0

= u

Simplify

- \frac{5}{2} - \frac{1}{2} : \frac{- 5 - 1}{2}

- \frac{5}{2} - \frac{1}{2}

Apply rule

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

= \frac{- 5 - 1}{2}

u \geq \frac{- 5 - 1}{2}

u \geq \frac{- 5 - 1}{2}

u \geq \frac{- 5 - 1}{2}

u + \frac{1}{2} \leq \frac{5}{4} : u \leq \frac{5 - 1}{2}

u + \frac{1}{2} \leq \frac{5}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

u + \frac{1}{2} \leq \frac{5}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u + \frac{1}{2} \leq \frac{5}{2}

Move

\frac{1}{2}

to the right side

u + \frac{1}{2} \leq \frac{5}{2}

Subtract

\frac{1}{2}

from both sides

u + \frac{1}{2} - \frac{1}{2} \leq \frac{5}{2} - \frac{1}{2}

Simplify

u + \frac{1}{2} - \frac{1}{2} \leq \frac{5}{2} - \frac{1}{2}

Simplify

u + \frac{1}{2} - \frac{1}{2} : u

u + \frac{1}{2} - \frac{1}{2}

Add similar elements:

\frac{1}{2} - \frac{1}{2} \leq 0

= u

Simplify

\frac{5}{2} - \frac{1}{2} : \frac{5 - 1}{2}

\frac{5}{2} - \frac{1}{2}

Apply rule

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

= \frac{5 - 1}{2}

u \leq \frac{5 - 1}{2}

u \leq \frac{5 - 1}{2}

u \leq \frac{5 - 1}{2}

Combine the intervals

u \geq \frac{- 5 - 1}{2} and u \leq \frac{5 - 1}{2}

Merge Overlapping Intervals

u \geq \frac{- 5 - 1}{2} and u \leq \frac{5 - 1}{2}

The intersection of two intervals is the set of numbers which are in both intervals

u \geq \frac{- 5 - 1}{2}

and

u \leq \frac{5 - 1}{2}

\frac{- 5 - 1}{2} \leq u \leq \frac{5 - 1}{2}

\frac{- 5 - 1}{2} \leq u \leq \frac{5 - 1}{2}

\frac{- 5 - 1}{2} \leq u \leq \frac{5 - 1}{2}

Substitute back

u = cos (x)

\frac{- 5 - 1}{2} \leq cos (x) \leq \frac{5 - 1}{2}

a \leq u \leq b

then

a \leq u and u \leq b

\frac{- 5 - 1}{2} \leq cos (x) and cos (x) \leq \frac{5 - 1}{2}

\frac{- 5 - 1}{2} \leq cos (x) :

True for all

x \in R

\frac{- 5 - 1}{2} \leq cos (x)

Switch sides

cos (x) \geq \frac{- 5 - 1}{2}

Range of

cos (x) : - 1 \leq cos (x) \leq 1

Function range definition

The range of the basic

cos

function is

- 1 \leq cos (x) \leq 1

- 1 \leq cos (x) \leq 1

cos (x) \geq \frac{- 5 - 1}{2} and - 1 \leq cos (x) \leq 1 : - 1 \leq cos (x) \leq 1

Let

y = cos (x)

Combine the intervals

y \geq \frac{- 5 - 1}{2} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \geq \frac{- 5 - 1}{2} and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y \geq \frac{- 5 - 1}{2}

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all x

True for all x \in R

cos (x) \leq \frac{5 - 1}{2} : arccos (\frac{5 - 1}{2}) + 2 πn \leq x \leq 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

cos (x) \leq \frac{5 - 1}{2}

For

cos (x) \leq a

, if

- 1 < a < 1

then

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

arccos (\frac{5 - 1}{2}) + 2 πn \leq x \leq 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

Combine the intervals

True for all x \in R and arccos (\frac{5 - 1}{2}) + 2 πn \leq x \leq 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

Merge Overlapping Intervals

arccos (\frac{5 - 1}{2}) + 2 πn \leq x \leq 2 π - arccos (\frac{5 - 1}{2}) + 2 πn

sin^{2} (x) \leq \frac{3}{2} sin (x) : 2 πn \leq x \leq \frac{π}{3} + 2 πn or \frac{2 π}{3} + 2 πn \leq x \leq π + 2 πn

sin^{2} (x) \leq \frac{3}{2} sin (x)

Let:

u = sin (x)

u^{2} \leq \frac{3}{2} u

u^{2} \leq \frac{3}{2} u : 0 \leq u \leq \frac{3}{2}

u^{2} \leq \frac{3}{2} u

Rewrite in standard form

u^{2} \leq \frac{3}{2} u

Subtract

\frac{3}{2} u

from both sides

u^{2} - \frac{3}{2} u \leq \frac{3}{2} u - \frac{3}{2} u

Simplify

u^{2} - \frac{3}{2} u \leq 0

Multiply both sides by

2

u^{2} \cdot 2 - \frac{3}{2} u \cdot 2 \leq 0 \cdot 2

2 u^{2} - 3 u \leq 0

2 u^{2} - 3 u \leq 0

Factor

2 u^{2} - 3 u : u (2 u - 3)

2 u^{2} - 3 u

Apply exponent rule:

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

u^{2} = uu

= 2 uu - 3 u

Factor out common term

u

Multiply the numbers:

1 \cdot 2 = 2

= u (2 u - 3)

u (2 u - 3) \leq 0

Identify the intervals

Find the signs of the factors of

u (2 u - 3)

Find the signs of

u

u = 0

u < 0

u > 0

Find the signs of

2 u - 3

2 u - 3 = 0 : u = \frac{3}{2}

2 u - 3 = 0

Move

3

to the right side

2 u - 3 = 0

Add

3

to both sides

2 u - 3 + 3 = 0 + 3

Simplify

2 u = 3

2 u = 3

Divide both sides by

2

2 u = 3

Divide both sides by

2

\frac{2 u}{2} = \frac{3}{2}

Simplify

u = \frac{3}{2}

u = \frac{3}{2}

2 u - 3 < 0 : u < \frac{3}{2}

2 u - 3 < 0

Move

3

to the right side

2 u - 3 < 0

Add

3

to both sides

2 u - 3 + 3 < 0 + 3

Simplify

2 u < 3

2 u < 3

Divide both sides by

2

2 u < 3

Divide both sides by

2

\frac{2 u}{2} < \frac{3}{2}

Simplify

u < \frac{3}{2}

u < \frac{3}{2}

2 u - 3 > 0 : u > \frac{3}{2}

2 u - 3 > 0

Move

3

to the right side

2 u - 3 > 0

Add

3

to both sides

2 u - 3 + 3 > 0 + 3

Simplify

2 u > 3

2 u > 3

Divide both sides by

2

2 u > 3

Divide both sides by

2

\frac{2 u}{2} > \frac{3}{2}

Simplify

u > \frac{3}{2}

u > \frac{3}{2}

Summarize in a table:

u 2 u - 3 u (2 u - 3) u < 0 - - + u = 0 0 - 0 0 < u < \frac{3}{2} + - - u = \frac{3}{2} + 00 u > \frac{3}{2} + + +

Identify the intervals that satisfy the required condition:

\leq 0

u = 0 or 0 < u < \frac{3}{2} or u = \frac{3}{2}

Merge Overlapping Intervals

0 \leq u < \frac{3}{2} or u = \frac{3}{2}

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

u = 0

0 < u < \frac{3}{2}

0 \leq u < \frac{3}{2}

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

0 \leq u < \frac{3}{2}

u = \frac{3}{2}

0 \leq u \leq \frac{3}{2}

0 \leq u \leq \frac{3}{2}

0 \leq u \leq \frac{3}{2}

0 \leq u \leq \frac{3}{2}

Substitute back

u = sin (x)

0 \leq sin (x) \leq \frac{3}{2}

a \leq u \leq b

then

a \leq u and u \leq b

0 \leq sin (x) and sin (x) \leq \frac{3}{2}

0 \leq sin (x) : 2 πn \leq x \leq π + 2 πn

0 \leq sin (x)

Switch sides

sin (x) \geq 0

For

sin (x) \geq a

, if

- 1 < a < 1

then

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

arcsin (0) + 2 πn \leq x \leq π - arcsin (0) + 2 πn

Simplify

arcsin (0) : 0

arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

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}

= 0

Simplify

π - arcsin (0) : π

π - arcsin (0)

Use the following trivial identity:

arcsin (0) = 0

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 - 0

π - 0 = π

= π

0 + 2 πn \leq x \leq π + 2 πn

Simplify

2 πn \leq x \leq π + 2 πn

sin (x) \leq \frac{3}{2} : - \frac{4 π}{3} + 2 πn \leq x \leq \frac{π}{3} + 2 πn

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

For

sin (x) \leq a

, if

- 1 < a < 1

then

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

- π - arcsin (\frac{3}{2}) + 2 πn \leq x \leq arcsin (\frac{3}{2}) + 2 πn

Simplify

- π - arcsin (\frac{3}{2}) : - \frac{4 π}{3}

- π - arcsin (\frac{3}{2})

Use the following trivial identity:

arcsin (\frac{3}{2}) = \frac{π}{3}

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}{3}

Simplify

- π - \frac{π}{3}

Convert element to fraction:

π = \frac{π 3}{3}

= - \frac{π 3}{3} - \frac{π}{3}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 3 - π}{3}

Add similar elements:

- 3 π - π = - 4 π

= \frac{- 4 π}{3}

Apply the fraction rule:

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

= - \frac{4 π}{3}

= - \frac{4 π}{3}

Simplify

arcsin (\frac{3}{2}) : \frac{π}{3}

arcsin (\frac{3}{2})

Use the following trivial identity:

arcsin (\frac{3}{2}) = \frac{π}{3}

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}{3}

- \frac{4 π}{3} + 2 πn \leq x \leq \frac{π}{3} + 2 πn

Combine the intervals

2 πn \leq x \leq π + 2 πn and - \frac{4 π}{3} + 2 πn \leq x \leq \frac{π}{3} + 2 πn

Merge Overlapping Intervals

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

Combine the intervals

arccos (\frac{5 - 1}{2}) + 2 πn \leq x \leq 2 π - arccos (\frac{5 - 1}{2}) + 2 πn and (2 πn \leq x \leq \frac{π}{3} + 2 πn or \frac{2 π}{3} + 2 πn \leq x \leq π + 2 πn)

Merge Overlapping Intervals

arccos (\frac{5 - 1}{2}) + 2 πn \leq x \leq \frac{π}{3} + 2 πn or \frac{2 π}{3} + 2 πn \leq x \leq π + 2 πn

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