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Popular >

cos^2(x)>= cos(x)

Solution

cos^{2} (x) \geq cos (x)

Solution

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

Interval Notation

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

Decimal Notation

1.57079 \dots + 2 πn \leq x \leq 4.71238 \dots + 2 πn

Solution steps

cos^{2} (x) \geq cos (x)

Let:

u = cos (x)

u^{2} \geq u

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

u^{2} \geq u

Rewrite in standard form

u^{2} \geq u

Subtract

u

from both sides

u^{2} - u \geq u - u

Simplify

u^{2} - u \geq 0

u^{2} - u \geq 0

Factor

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

u^{2} - u

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) \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 = cos (x)

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

cos (x) \leq 0 : \frac{π}{2} + 2 πn \leq x \leq \frac{3 π}{2} + 2 πn

cos (x) \leq 0

For

cos (x) \leq a

, if

- 1 < a < 1

then

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

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

Simplify

arccos (0) : \frac{π}{2}

arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2}

Simplify

2 π - arccos (0) : \frac{3 π}{2}

2 π - arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{2}

Simplify

2 π - \frac{π}{2}

Convert element to fraction:

2 π = \frac{2 π 2}{2}

= \frac{2 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{2 π 2 - π}{2}

2 π 2 - π = 3 π

2 π 2 - π

Multiply the numbers:

2 \cdot 2 = 4

= 4 π - π

Add similar elements:

4 π - π = 3 π

= 3 π

= \frac{3 π}{2}

= \frac{3 π}{2}

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

cos (x) \geq 1 :

No Solution for

x \in R

cos (x) \geq 1

For

cos (x) \geq a

, if

- 1 < a < 1

then

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

- arccos (1) + 2 πn \leq x \leq arccos (1) + 2 πn

Simplify

- arccos (1) : 0

- arccos (1)

Use the following trivial identity:

arccos (1) = 0

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - 0

= 0

Simplify

arccos (1) : 0

arccos (1)

Use the following trivial identity:

arccos (1) = 0

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 0

0 + 2 πn \leq x \leq 0 + 2 πn

Simplify

No Solution for x \in R

Combine the intervals

\frac{π}{2} + 2 πn \leq x \leq \frac{3 π}{2} + 2 πn or False for all x \in R

Merge Overlapping Intervals

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

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