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Popular Trigonometry >

cos^2(x)+sin(x)+1>= 0

Solution

cos^{2} (x) + sin (x) + 1 \geq 0

Solution

True for all x \in R

Interval Notation

(- \infty, \infty)

Solution steps

cos^{2} (x) + sin (x) + 1 \geq 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) + 1 \geq 0

Simplify

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

Let:

u = sin (x)

u - u^{2} + 2 \geq 0

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

u - u^{2} + 2 \geq 0

Factor

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

u - u^{2} + 2

Factor out common term

- 1

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

Factor

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

u^{2} - u - 2

Write in the standard form

a x^{2} + b x + c

= u^{2} - u - 2

Break the expression into groups

u^{2} - u - 2

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

TrueCheck

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

False

u = 1, v = - 2

Group into

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

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

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

Factor out

u

from

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)

Factor out

- 2

from

- 2 u - 2 : - 2 (u + 1)

- 2 u - 2

Factor out common term

- 2

= - 2 (u + 1)

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

Factor out common term

u + 1

= (u + 1) (u - 2)

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

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

Identify the intervals

Find the signs of the factors of

(u + 1) (u - 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

Find the signs of

u - 2

u - 2 = 0 : u = 2

u - 2 = 0

Move

2

to the right side

u - 2 = 0

Add

2

to both sides

u - 2 + 2 = 0 + 2

Simplify

u = 2

u = 2

u - 2 < 0 : u < 2

u - 2 < 0

Move

2

to the right side

u - 2 < 0

Add

2

to both sides

u - 2 + 2 < 0 + 2

Simplify

u < 2

u < 2

u - 2 > 0 : u > 2

u - 2 > 0

Move

2

to the right side

u - 2 > 0

Add

2

to both sides

u - 2 + 2 > 0 + 2

Simplify

u > 2

u > 2

Summarize in a table:

u + 1 u - 2 (u + 1) (u - 2) u < - 1 - - + u = - 1 0 - 0 - 1 < u < 2 + - - u = 2 + 00 u > 2 + + +

Identify the intervals that satisfy the required condition:

\leq 0

u = - 1 or - 1 < u < 2 or u = 2

Merge Overlapping Intervals

- 1 \leq u < 2 or u = 2

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

u = - 1

- 1 < u < 2

- 1 \leq u < 2

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

- 1 \leq u < 2

u = 2

- 1 \leq u \leq 2

- 1 \leq u \leq 2

- 1 \leq u \leq 2

- 1 \leq u \leq 2

Substitute back

u = sin (x)

- 1 \leq sin (x) \leq 2

a \leq u \leq b

then

a \leq u and u \leq b

- 1 \leq sin (x) and sin (x) \leq 2

- 1 \leq sin (x) :

True for all

x \in R

- 1 \leq sin (x)

Switch sides

sin (x) \geq - 1

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

Let

y = sin (x)

Combine the intervals

y \geq - 1 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \geq - 1 and - 1 \leq y \leq 1

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

y \geq - 1

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

sin (x) \leq 2 :

True for all

x \in R

sin (x) \leq 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) \leq 2 and - 1 \leq sin (x) \leq 1 : - 1 \leq sin (x) \leq 1

Let

y = sin (x)

Combine the intervals

y \leq 2 and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \leq 2 and - 1 \leq y \leq 1

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

y \leq 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

Combine the intervals

True for all x \in R and True for all x \in R

Merge Overlapping Intervals

True for all x \in R and True for all x \in R

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

True for all

x \in R

and

True for all

x \in R

True for all x \in R

True for all x

True for all x \in R

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