English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

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

Solution

2 + cos^{2} (x) + sin (x) > 0

Solution

True for all x \in R

Interval Notation

(- \infty, \infty)

Solution steps

2 + cos^{2} (x) + sin (x) > 0

Use the following identity:

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

Therefore

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

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

Simplify

sin (x) - sin^{2} (x) + 3 > 0

Let:

u = sin (x)

u - u^{2} + 3 > 0

u - u^{2} + 3 > 0 : \frac{- 13 + 1}{2} < u < \frac{13 + 1}{2}

u - u^{2} + 3 > 0

Complete the square

u - u^{2} + 3 : - (u - \frac{1}{2})^{2} + \frac{13}{4}

u - u^{2} + 3

Write in the standard form

a x^{2} + b x + c

- u^{2} + u + 3

Write

- u^{2} + u + 3

in the form:

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

Factor out

- 1

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

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 - 3 + (- \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} - 3 - (- \frac{1}{2})^{2})

Simplify

- (u - \frac{1}{2})^{2} + \frac{13}{4}

- (u - \frac{1}{2})^{2} + \frac{13}{4} > 0

Move

\frac{13}{4}

to the right side

- (u - \frac{1}{2})^{2} + \frac{13}{4} > 0

Subtract

\frac{13}{4}

from both sides

- (u - \frac{1}{2})^{2} + \frac{13}{4} - \frac{13}{4} > 0 - \frac{13}{4}

Simplify

- (u - \frac{1}{2})^{2} > - \frac{13}{4}

- (u - \frac{1}{2})^{2} > - \frac{13}{4}

Multiply both sides by

- 1

- (u - \frac{1}{2})^{2} > - \frac{13}{4}

Multiply both sides by -1 (reverse the inequality)

(- (u - \frac{1}{2})^{2}) (- 1) < (- \frac{13}{4}) (- 1)

Simplify

(u - \frac{1}{2})^{2} < \frac{13}{4}

(u - \frac{1}{2})^{2} < \frac{13}{4}

For

u^{n} < a

, if

n

is even then

- \frac{13}{4} < u - \frac{1}{2} < \frac{13}{4}

a < u < b

then

a < u and u < b

- \frac{13}{4} < u - \frac{1}{2} and u - \frac{1}{2} < \frac{13}{4}

- \frac{13}{4} < u - \frac{1}{2} : u > \frac{- 13 + 1}{2}

- \frac{13}{4} < u - \frac{1}{2}

Switch sides

u - \frac{1}{2} > - \frac{13}{4}

Simplify

\frac{13}{4} : \frac{13}{2}

\frac{13}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{13}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{13}{2}

u - \frac{1}{2} > - \frac{13}{2}

Move

\frac{1}{2}

to the right side

u - \frac{1}{2} > - \frac{13}{2}

Add

\frac{1}{2}

to both sides

u - \frac{1}{2} + \frac{1}{2} > - \frac{13}{2} + \frac{1}{2}

Simplify

u - \frac{1}{2} + \frac{1}{2} > - \frac{13}{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} > 0

= u

Simplify

- \frac{13}{2} + \frac{1}{2} : \frac{- 13 + 1}{2}

- \frac{13}{2} + \frac{1}{2}

Apply rule

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

= \frac{- 13 + 1}{2}

u > \frac{- 13 + 1}{2}

u > \frac{- 13 + 1}{2}

u > \frac{- 13 + 1}{2}

u - \frac{1}{2} < \frac{13}{4} : u < \frac{13 + 1}{2}

u - \frac{1}{2} < \frac{13}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

u - \frac{1}{2} < \frac{13}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

u - \frac{1}{2} < \frac{13}{2}

Move

\frac{1}{2}

to the right side

u - \frac{1}{2} < \frac{13}{2}

Add

\frac{1}{2}

to both sides

u - \frac{1}{2} + \frac{1}{2} < \frac{13}{2} + \frac{1}{2}

Simplify

u - \frac{1}{2} + \frac{1}{2} < \frac{13}{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} < 0

= u

Simplify

\frac{13}{2} + \frac{1}{2} : \frac{13 + 1}{2}

\frac{13}{2} + \frac{1}{2}

Apply rule

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

= \frac{13 + 1}{2}

u < \frac{13 + 1}{2}

u < \frac{13 + 1}{2}

u < \frac{13 + 1}{2}

Combine the intervals

u > \frac{- 13 + 1}{2} and u < \frac{13 + 1}{2}

Merge Overlapping Intervals

u > \frac{- 13 + 1}{2} and u < \frac{13 + 1}{2}

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

u > \frac{- 13 + 1}{2}

and

u < \frac{13 + 1}{2}

\frac{- 13 + 1}{2} < u < \frac{13 + 1}{2}

\frac{- 13 + 1}{2} < u < \frac{13 + 1}{2}

\frac{- 13 + 1}{2} < u < \frac{13 + 1}{2}

Substitute back

u = sin (x)

\frac{- 13 + 1}{2} < sin (x) < \frac{13 + 1}{2}

a < u < b

then

a < u and u < b

\frac{- 13 + 1}{2} < sin (x) and sin (x) < \frac{13 + 1}{2}

\frac{- 13 + 1}{2} < sin (x) :

True for all

x \in R

\frac{- 13 + 1}{2} < sin (x)

Switch sides

sin (x) > \frac{- 13 + 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) > \frac{- 13 + 1}{2} and - 1 \leq sin (x) \leq 1 : - 1 \leq sin (x) \leq 1

Let

y = sin (x)

Combine the intervals

y > \frac{- 13 + 1}{2} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y > \frac{- 13 + 1}{2} and - 1 \leq y \leq 1

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

y > \frac{- 13 + 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

sin (x) < \frac{13 + 1}{2} :

True for all

x \in R

sin (x) < \frac{13 + 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) < \frac{13 + 1}{2} and - 1 \leq sin (x) \leq 1 : - 1 \leq sin (x) \leq 1

Let

y = sin (x)

Combine the intervals

y < \frac{13 + 1}{2} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y < \frac{13 + 1}{2} and - 1 \leq y \leq 1

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

y < \frac{13 + 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

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

Popular Examples

2sin^2(x)-1>0

10<5sin(pi/6 (x+2))+6

arctan(1-|x|)<= 1/2

sin(t)< 1/3 ln(1)

5-8cos(x)+4cos(2x)<0