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Popular >

sin^2(x)<= 1

Solution

sin^{2} (x) \leq 1

Solution

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

sin^{2} (x) \leq 1

For

u^{n} \leq a

, if

n

is even then

- 1 \leq sin (x) \leq 1

a \leq u \leq b

then

a \leq u and u \leq b

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

- 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 1 :

True for all

x \in R

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

Let

y = sin (x)

Combine the intervals

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

Merge Overlapping Intervals

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

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

y \leq 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

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

Popular Examples

cot(x)+(sin(x))/(cos(x)-2)>= 0

cot (x) + \frac{sin ( x )}{cos ( x ) - 2} \geq 0

(2sin(2x)+sqrt(2))*tan(x)<0

(2 sin (2 x) + 2) \cdot tan (x) < 0

cos(x)<=-(sqrt(2))/2 ,-pi<= x<= pi

cos (x) \leq - \frac{2}{2}, - π \leq x \leq π

6sin(2x-(2pi)/3)>0

6 sin (2 x - \frac{2 π}{3}) > 0

1>tan(x)

1 > tan (x)