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Popular >

-pi/2 <arcsin(x)< pi/2

Solution

- \frac{π}{2} < arcsin (x) < \frac{π}{2}

Solution

- 1 < x < 1

Interval Notation

(- 1, 1)

Solution steps

- \frac{π}{2} < arcsin (x) < \frac{π}{2}

a < u < b

then

a < u and u < b

- \frac{π}{2} < arcsin (x) and arcsin (x) < \frac{π}{2}

- \frac{π}{2} < arcsin (x) : - 1 < x \leq 1

- \frac{π}{2} < arcsin (x)

Switch sides

arcsin (x) > - \frac{π}{2}

arcsin (x) > a

then

x > sin (a)

x > sin (- \frac{π}{2})

sin (- \frac{π}{2}) = - 1

sin (- \frac{π}{2})

Use the following property:

sin (- x) = - sin (x)

sin (- \frac{π}{2}) = - sin (\frac{π}{2})

= - sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (\frac{π}{2})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 1

= - 1

x > - 1

Domain of

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

Domain definition

Find known functions domain restrictions:

- 1 \leq x \leq 1

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

Solve

- 1 \leq x \leq 1 : - 1 \leq x \leq 1

- 1 \leq x \leq 1

Could not simplify further

- 1 \leq x \leq 1

The function domain

- 1 \leq x \leq 1

Combine the intervals

x > - 1 and - 1 \leq x \leq 1

Merge Overlapping Intervals

x > - 1 and - 1 \leq x \leq 1

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

x > - 1

and

- 1 \leq x \leq 1

- 1 < x \leq 1

- 1 < x \leq 1

arcsin (x) < \frac{π}{2} : - 1 \leq x < 1

arcsin (x) < \frac{π}{2}

arcsin (x) < a

then

x < sin (a)

x < sin (\frac{π}{2})

sin (\frac{π}{2}) = 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 1

x < 1

Domain of

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

Domain definition

Find known functions domain restrictions:

- 1 \leq x \leq 1

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

Solve

- 1 \leq x \leq 1 : - 1 \leq x \leq 1

- 1 \leq x \leq 1

Could not simplify further

- 1 \leq x \leq 1

The function domain

- 1 \leq x \leq 1

Combine the intervals

x < 1 and - 1 \leq x \leq 1

Merge Overlapping Intervals

x < 1 and - 1 \leq x \leq 1

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

x < 1

and

- 1 \leq x \leq 1

- 1 \leq x < 1

- 1 \leq x < 1

Combine the intervals

- 1 < x \leq 1 and - 1 \leq x < 1

Merge Overlapping Intervals

- 1 < x \leq 1 and - 1 \leq x < 1

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

- 1 < x \leq 1

and

- 1 \leq x < 1

- 1 < x < 1

- 1 < x < 1

Popular Examples

-sqrt(2)<= cos(2x)<= sqrt(2)

- 2 \leq cos (2 x) \leq 2

(11pi)/9 <= arctan(θ)<= (13pi)/9

\frac{11 π}{9} \leq arctan (θ) \leq \frac{13 π}{9}

sin(x)=-(sqrt(3))/5 \land cos(x)>0

sin (x) = - \frac{3}{5} and cos (x) > 0

cos(θ)=0.222\land tan(θ)<0,sin(θ)

cos (θ) = 0.222 and tan (θ) < 0, sin (θ)

sin(x)0<= x<= pi

sin (x) 0 \leq x \leq π