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Popular Trigonometry >

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

Solution

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

Solution

x = π + 2 πn

Decimal

x = 3.14159 \dots + 2 πn

Solution steps

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

a \leq u \leq b

then

a \leq u and u \leq b

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

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

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

Let:

u = sin (x)

u \leq u^{2}

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

u \leq u^{2}

Rewrite in standard form

u \leq u^{2}

Subtract

u^{2}

from both sides

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

Simplify

u - u^{2} \leq 0

u - u^{2} \leq 0

Factor

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

u - u^{2}

Apply exponent rule:

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

u^{2} = uu

= - uu + u

Factor out common term

- u

= - u (u - 1)

- u (u - 1) \leq 0

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

u (u - 1) \geq 0

Identify the intervals

Find the signs of the factors of

u (u - 1)

Find the signs of

u

u = 0

u < 0

u > 0

Find the signs of

u - 1

u - 1 = 0 : u = 1

u - 1 = 0

Move

1

to the right side

u - 1 = 0

Add

1

to 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

Add

1

to 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

Add

1

to both sides

u - 1 + 1 > 0 + 1

Simplify

u > 1

u > 1

Summarize in a table:

u u - 1 u (u - 1) u < 0 - - + u = 0 0 - 0 0 < u < 1 + - - u = 1 + 00 u > 1 + + +

Identify the intervals that satisfy the required condition:

\geq 0

u < 0 or u = 0 or u = 1 or u > 1

Merge Overlapping Intervals

u \leq 0 or u = 1 or u > 1

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

u < 0

u = 0

u \leq 0

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

u \leq 0

u = 1

u \leq 0 or u = 1

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

u \leq 0 or u = 1

u > 1

u \leq 0 or u \geq 1

u \leq 0 or u \geq 1

u \leq 0 or u \geq 1

u \leq 0 or u \geq 1

Substitute back

u = sin (x)

sin (x) \leq 0 or sin (x) \geq 1

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

sin (x) \leq 0

For

sin (x) \leq a

, if

- 1 < a < 1

then

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

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

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

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

- π + 2 πn \leq x \leq 0 + 2 πn

Simplify

- π + 2 πn \leq x \leq 2 πn

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

sin (x) \geq 1

For

sin (x) \geq a

, if

- 1 < a < 1

then

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

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

Simplify

arcsin (1) : \frac{π}{2}

arcsin (1)

Use the following trivial identity:

arcsin (1) = \frac{π}{2}

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}

Simplify

π - arcsin (1) : \frac{π}{2}

π - arcsin (1)

Use the following trivial identity:

arcsin (1) = \frac{π}{2}

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}{2}

Simplify

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}

= \frac{π}{2}

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

Simplify

x = \frac{π}{2} + 2 πn

Combine the intervals

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

Merge Overlapping Intervals

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

= u (2 u - 1 \cdot 2 3)

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

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

x = π + 2 πn

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