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Popular >

cos^2(x)<=-sin(x)

Solution

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

Solution

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

Interval Notation

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

Decimal Notation

- 2.47535 \dots + 2 πn \leq x \leq - 0.66623 \dots + 2 πn

Solution steps

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

Move

sin (x)

to the left side

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

Add

sin (x)

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

Therefore

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

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

Let:

u = sin (x)

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

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

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

Complete the square

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

1 - u^{2} + u

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}

Factor out

- 1

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

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

Subtract

\frac{5}{4}

from 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}

Multiply both sides by

- 1

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

Multiply both sides by -1 (reverse the inequality)

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

Simplify

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

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

For

u^{n} \geq a

, if

n

is even then

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

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

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

Move

\frac{1}{2}

to the right side

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

Add

\frac{1}{2}

to 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}

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

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

Apply radical rule:

assuming

a \geq 0, b \geq 0

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

Move

\frac{1}{2}

to the right side

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

Add

\frac{1}{2}

to 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}

Combine the intervals

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

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

Substitute back

u = sin (x)

sin (x) \leq \frac{- 5 + 1}{2} or sin (x) \geq \frac{5 + 1}{2}

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

sin (x) \leq \frac{- 5 + 1}{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{- 5 + 1}{2}) + 2 πn \leq x \leq arcsin (\frac{- 5 + 1}{2}) + 2 πn

sin (x) \geq \frac{5 + 1}{2} :

False for all

x \in R

sin (x) \geq \frac{5 + 1}{2}

Range of

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

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (x) \leq 1

- 1 \leq sin (x) \leq 1

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

False

Let

y = sin (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

False for all y \in R

False for all y \in R

No Solution for x \in R

False for all x \in R

Combine the intervals

- π - arcsin (\frac{- 5 + 1}{2}) + 2 πn \leq x \leq arcsin (\frac{- 5 + 1}{2}) + 2 πn or False for all x \in R

Merge Overlapping Intervals

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

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