English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Trigonometry >

-1<= arccos(x^2)<= 1

Solution

- 1 \leq arccos (x^{2}) \leq 1

Solution

- 1 \leq x \leq - cos (1) or cos (1) \leq x \leq 1

Interval Notation

[- 1, - cos (1)] \cup [cos (1), 1]

Decimal

- 1 \leq x \leq - 0.73505 \dots or 0.73505 \dots \leq x \leq 1

Solution steps

- 1 \leq arccos (x^{2}) \leq 1

a \leq u \leq b

then

a \leq u and u \leq b

- 1 \leq arccos (x^{2}) and arccos (x^{2}) \leq 1

- 1 \leq arccos (x^{2}) : - 1 \leq x \leq 1

- 1 \leq arccos (x^{2})

Switch sides

arccos (x^{2}) \geq - 1

Range of

arccos (x^{2}) : 0 \leq arccos (x^{2}) \leq \frac{π}{2}

Function range definition

Range of

x^{2} : f (x) \geq 0

Function range definition

Vertex of

x^{2} :

Minimum

(0, 0)

Parabola equation in polynomial form

The parabola parameters are:

a = 1, b = 0, c = 0

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{0}{2 \cdot 1}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 0^{2}

Simplify

y_{v} = 0

y_{v} = 0

Therefore the parabola vertex is

(0, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, 0)

For a parabola

a x^{2} + b x + c

with Vertex

(x_{v}, y_{v})

a < 0

the range is

f (x) \leq y_{v}

a > 0

the range is

f (x) \geq y_{v}

a = 1,

Vertex

(x_{v}, y_{v}) = (0, 0)

f (x) \geq 0

Since

arccos

is a decreasing function with range of

0 \leq arccos (x) \leq π

and

x^{2} \geq 0

0 \leq arccos (x^{2}) \leq arccos (0)

Simplify

0 \leq arccos (x^{2}) \leq \frac{π}{2}

arccos (x^{2}) \geq - 1 and 0 \leq arccos (x^{2}) \leq \frac{π}{2} : 0 \leq arccos (x^{2}) \leq \frac{π}{2}

Let

y = arccos (x^{2})

Combine the intervals

y \geq - 1 and 0 \leq y \leq \frac{π}{2}

Merge Overlapping Intervals

y \geq - 1 and 0 \leq y \leq \frac{π}{2}

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

y \geq - 1

and

0 \leq y \leq \frac{π}{2}

0 \leq y \leq \frac{π}{2}

0 \leq y \leq \frac{π}{2}

True for all x in domain of arccos (x^{2})

Domain of

arccos (x^{2}) : - 1 \leq x \leq 1

Domain definition

Find known functions domain restrictions:

- 1 \leq x \leq 1

arccos (f (x)) \Rightarrow - 1 \leq f (x) \leq 1

Solve

- 1 \leq x^{2} \leq 1 : - 1 \leq x \leq 1

- 1 \leq x^{2} \leq 1

a \leq u \leq b

then

a \leq u and u \leq b

- 1 \leq x^{2} and x^{2} \leq 1

- 1 \leq x^{2} :

True for all

x \in R

- 1 \leq x^{2}

Switch sides

x^{2} \geq - 1

If n is even,

u^{n} \geq 0

for all

u

True for all x

x^{2} \leq 1 : - 1 \leq x \leq 1

x^{2} \leq 1

For

u^{n} \leq a

, if

n

is even then

- 1 \leq x \leq 1

Combine the intervals

True for all x \in R and - 1 \leq x \leq 1

Merge Overlapping Intervals

True for all x \in R and - 1 \leq x \leq 1

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

True for all

x \in R

and

- 1 \leq x \leq 1

- 1 \leq x \leq 1

- 1 \leq x \leq 1

The function domain

- 1 \leq x \leq 1

- 1 \leq x \leq 1

arccos (x^{2}) \leq 1 : - 1 \leq x \leq - cos (1) or cos (1) \leq x \leq 1

arccos (x^{2}) \leq 1

arccos (x) \leq a

then

x \geq cos (a)

x^{2} \geq cos (1)

x^{2} \geq cos (1) : x \leq - cos (1) or x \geq cos (1)

x^{2} \geq cos (1)

For

u^{n} \geq a

, if

n

is even then

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

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

Domain of

arccos (x^{2}) : - 1 \leq x \leq 1

Domain definition

Find known functions domain restrictions:

- 1 \leq x \leq 1

arccos (f (x)) \Rightarrow - 1 \leq f (x) \leq 1

Solve

- 1 \leq x^{2} \leq 1 : - 1 \leq x \leq 1

- 1 \leq x^{2} \leq 1

a \leq u \leq b

then

a \leq u and u \leq b

- 1 \leq x^{2} and x^{2} \leq 1

- 1 \leq x^{2} :

True for all

x \in R

- 1 \leq x^{2}

Switch sides

x^{2} \geq - 1

If n is even,

u^{n} \geq 0

for all

u

True for all x

x^{2} \leq 1 : - 1 \leq x \leq 1

x^{2} \leq 1

For

u^{n} \leq a

, if

n

is even then

- 1 \leq x \leq 1

Combine the intervals

True for all x \in R and - 1 \leq x \leq 1

Merge Overlapping Intervals

True for all x \in R and - 1 \leq x \leq 1

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

True for all

x \in R

and

- 1 \leq x \leq 1

- 1 \leq x \leq 1

- 1 \leq x \leq 1

The function domain

- 1 \leq x \leq 1

Combine the intervals

x \leq - cos (1) or x \geq cos (1) and - 1 \leq x \leq 1

Merge Overlapping Intervals

x \leq - cos (1) or x \geq cos (1) and - 1 \leq x \leq 1

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

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

and

- 1 \leq x \leq 1

- 1 \leq x \leq - cos (1) or cos (1) \leq x \leq 1

- 1 \leq x \leq - cos (1) or cos (1) \leq x \leq 1

Combine the intervals

- 1 \leq x \leq 1 and (- 1 \leq x \leq - cos (1) or cos (1) \leq x \leq 1)

Merge Overlapping Intervals

- 1 \leq x \leq 1 and - 1 \leq x \leq - cos (1) or cos (1) \leq x \leq 1

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

- 1 \leq x \leq 1

and

- 1 \leq x \leq - cos (1) or cos (1) \leq x \leq 1

- 1 \leq x \leq - cos (1) or cos (1) \leq x \leq 1

- 1 \leq x \leq - cos (1) or cos (1) \leq x \leq 1

Graph

Popular Examples

1-cos(θ)0<= θ<= 2pi

4(1-sin(θ))0<= θ<= pi

-1<sin(x)<-1/2

sin(θ)>0\land tan(θ)>0

tan(x)<0<5sin(x)