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Popular Trigonometry >

9^{1+sin^2(pix)}+30*9^{cos^2(pix)}<= 117

Solution

9^{1 + s i n^{2} (π x)} + 30 \cdot 9^{c o s^{2} (π x)} \leq 117

Solution

False for all x \in R

Solution steps

9^{1 + s i n^{2} (π x)} + 30 \cdot 9^{c o s^{2} (π x)} \leq 117

Use the following identity:

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

Therefore

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

9^{1 + s i n^{2} (π x)} + 30 \cdot 9^{1 - s i n^{2} (π x)} \leq 117

Simplify

9^{1 + s i n^{2} (π x)} + 30 \cdot 9^{1 - s i n^{2} (π x)} : 9^{1 + s i n^{2} (π x)} + 10 \cdot 3^{- 2 s i n^{2} (π x) + 3}

9^{1 + s i n^{2} (π x)} + 30 \cdot 9^{1 - s i n^{2} (π x)}

30 \cdot 9^{1 - s i n^{2} (π x)} = 10 \cdot 3^{- 2 s i n^{2} (π x) + 3}

30 \cdot 9^{1 - s i n^{2} (π x)}

Factor integer

30 = 3 \cdot 10

= 3 \cdot 10 \cdot 9^{1 - s i n^{2} (π x)}

Factor integer

9 = 3^{2}

= 3 \cdot 10 (3^{2})^{1 - s i n^{2} (π x)}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

(3^{2})^{1 - s i n^{2} (π x)} = 3^{2 (1 - s i n^{2} (π x))}

= 3 \cdot 10 \cdot 3^{2 (1 - s i n^{2} (π x))}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

3 \cdot 3^{2 (1 - s i n^{2} (π x))} = 3^{1 + 2 (- s i n^{2} (π x) + 1)}

= 10 \cdot 3^{1 + 2 (1 - s i n^{2} (π x))}

3^{1 + 2 (1 - s i n^{2} (π x))} = 3^{- 2 s i n^{2} (π x) + 3}

3^{1 + 2 (1 - s i n^{2} (π x))}

Expand

1 + 2 (1 - sin^{2} (π x)) : - 2 sin^{2} (π x) + 3

1 + 2 (1 - sin^{2} (π x))

Expand

2 (1 - sin^{2} (π x)) : 2 - 2 sin^{2} (π x)

2 (1 - sin^{2} (π x))

Apply the distributive law:

a (b - c) = ab - a c

a = 2, b = 1, c = sin^{2} (π x)

= 2 \cdot 1 - 2 sin^{2} (π x)

Multiply the numbers:

2 \cdot 1 = 2

= 2 - 2 sin^{2} (π x)

= 1 + 2 - 2 sin^{2} (π x)

Add the numbers:

1 + 2 = 3

= - 2 sin^{2} (π x) + 3

= 3^{- 2 s i n^{2} (π x) + 3}

= 10 \cdot 3^{- 2 s i n^{2} (π x) + 3}

= 9^{s i n^{2} (π x) + 1} + 10 \cdot 3^{- 2 s i n^{2} (π x) + 3}

9^{1 + s i n^{2} (π x)} + 10 \cdot 3^{- 2 s i n^{2} (π x) + 3} \leq 117

Apply exponent rules

9^{1 + s i n^{2} (π x)} + 10 \cdot 3^{- 2 s i n^{2} (π x) + 3} \leq 117

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

9^{1 + s i n^{2} (π x)} = 9 \cdot 9^{s i n^{2} (π x)}

9 \cdot 9^{s i n^{2} (π x)} + 10 \cdot 3^{- 2 s i n^{2} (π x) + 3} \leq 117

Apply exponent rule:

a^{b - c} = \frac{a ^{b}}{a ^{c}}

3^{- 2 s i n^{2} (π x) + 3} = \frac{3 ^{3}}{3 ^{2 s i n^{2} (π x)}}

9 \cdot 9^{s i n^{2} (π x)} + 10 \cdot \frac{3 ^{3}}{3 ^{2 s i n^{2} (π x)}} \leq 117

f (x) > 0

we can multiply or devide both sides of inequality by

f (x)

3^{2 s i n^{2} (π x)}

is greater than

0

for all

x

9 \cdot 9^{s i n^{2} (π x)} \cdot 3^{2 s i n^{2} (π x)} + 10 \cdot \frac{3 ^{3}}{3 ^{2 s i n^{2} (π x)}} \cdot 3^{2 s i n^{2} (π x)} \leq 117 \cdot 3^{2 s i n^{2} (π x)}

Simplify

9 \cdot 9^{s i n^{2} (π x)} \cdot 3^{2 s i n^{2} (π x)} + 270 \leq 117 \cdot 3^{2 s i n^{2} (π x)}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

3^{2 s i n^{2} (π x)} = (3^{s i n^{2} (π x)})^{2}

9 \cdot 9^{s i n^{2} (π x)} (3^{s i n^{2} (π x)})^{2} + 270 \leq 117 (3^{s i n^{2} (π x)})^{2}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

3^{2 s i n^{2} (π x)} = (3^{s i n^{2} (π x)})^{2}

9 \cdot 9^{s i n^{2} (π x)} (3^{s i n^{2} (π x)})^{2} + 270 \leq 117 (3^{s i n^{2} (π x)})^{2}

Rewrite

9^{s i n^{2} (π x)}

3^{2 s i n^{2} (π x)}

9^{s i n^{2} (π x)}

9 = 3^{2}

= (3^{2})^{s i n^{2} (π x)}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

(3^{2})^{s i n^{2} (π x)} = 3^{2 s i n^{2} (π x)}

= 3^{2 s i n^{2} (π x)}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

3^{2 s i n^{2} (π x)} = (3^{s i n^{2} (π x)})^{2}

9 (3^{s i n^{2} (π x)})^{2} (3^{s i n^{2} (π x)})^{2} + 270 \leq 117 (3^{s i n^{2} (π x)})^{2}

9 (3^{s i n^{2} (π x)})^{2} (3^{s i n^{2} (π x)})^{2} + 270 \leq 117 (3^{s i n^{2} (π x)})^{2}

Let

v = 3^{s i n^{2} (π x)}

9 v^{2} v^{2} + 270 \leq 117 v^{2}

9 v^{2} v^{2} + 270 \leq 117 v^{2} : - 10 \leq v \leq - 3 or 3 \leq v \leq 10

9 v^{2} v^{2} + 270 \leq 117 v^{2}

Rewrite in standard form

9 v^{2} v^{2} + 270 \leq 117 v^{2}

Simplify

9 v^{2} v^{2} + 270 : 9 v^{4} + 270

9 v^{2} v^{2} + 270

9 v^{2} v^{2} = 9 v^{4}

9 v^{2} v^{2}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

v^{2} v^{2} = v^{2 + 2}

= 9 v^{2 + 2}

Add the numbers:

2 + 2 = 4

= 9 v^{4}

= 9 v^{4} + 270

9 v^{4} + 270 \leq 117 v^{2}

Subtract

117 v^{2}

from both sides

9 v^{4} + 270 - 117 v^{2} \leq 117 v^{2} - 117 v^{2}

Simplify

9 v^{4} + 270 - 117 v^{2} \leq 0

Divide both sides by

9

\frac{9 v ^{4}}{9} + \frac{270}{9} - \frac{117 v ^{2}}{9} \leq \frac{0}{9}

Refine

\frac{9 v ^{4}}{9} + \frac{270}{9} - \frac{117 v ^{2}}{9} \leq \frac{0}{9} : v^{4} - 13 v^{2} + 30 \leq 0

\frac{9 v ^{4}}{9} + \frac{270}{9} - \frac{117 v ^{2}}{9} \leq \frac{0}{9}

Simplify

\frac{9 v ^{4}}{9} + \frac{270}{9} - \frac{117 v ^{2}}{9} : v^{4} - 13 v^{2} + 30

\frac{9 v ^{4}}{9} + \frac{270}{9} - \frac{117 v ^{2}}{9}

Divide the numbers:

\frac{9}{9} = 1

= v^{4} + \frac{270}{9} - \frac{117 v ^{2}}{9}

Divide the numbers:

\frac{270}{9} = 30

= v^{4} + 30 - \frac{117 v ^{2}}{9}

Divide the numbers:

\frac{117}{9} = 13

= v^{4} + 30 - 13 v^{2}

Rewrite in standard form

= v^{4} - 13 v^{2} + 30

\frac{0}{9} = 0

\frac{0}{9}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

v^{4} - 13 v^{2} + 30 \leq 0

v^{4} - 13 v^{2} + 30 \leq 0

v^{4} - 13 v^{2} + 30 \leq 0

Factor

v^{4} - 13 v^{2} + 30 : (v + 3) (v - 3) (v + 10) (v - 10)

v^{4} - 13 v^{2} + 30

Let

u = v^{2}

= u^{2} - 13 u + 30

Factor

u^{2} - 13 u + 30 : (u - 3) (u - 10)

u^{2} - 13 u + 30

Break the expression into groups

u^{2} - 13 u + 30

Definition

Factors of

30 : 1, 2, 3, 5, 6, 10, 15, 30

30

Divisors (Factors)

Find the Prime factors of

30 : 2, 3, 5

30

30

divides by

230 = 15 \cdot 2

= 2 \cdot 15

15

divides by

315 = 5 \cdot 3

= 2 \cdot 3 \cdot 5

2, 3, 5

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3 \cdot 5

Multiply the prime factors of

30 : 6, 10, 15

2 \cdot 3 = 6

2 \cdot 5 = 10

6, 10, 15

6, 10, 15

Add the prime factors:

2, 3, 5

Add 1 and the number

30

itself

1, 30

The factors of

30

1, 2, 3, 5, 6, 10, 15, 30

Negative factors of

30 : - 1, - 2, - 3, - 5, - 6, - 10, - 15, - 30

Multiply the factors by

- 1

to get the negative factors

- 1, - 2, - 3, - 5, - 6, - 10, - 15, - 30

For every two factors such that

u * v = 30,

check if

u + v = - 13

Check

u = 1, v = 30 : u * v = 30, u + v = 31 \Rightarrow

FalseCheck

u = 2, v = 15 : u * v = 30, u + v = 17 \Rightarrow

False

u = - 3, v = - 10

Group into

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

(u^{2} - 3 u) + (- 10 u + 30)

= (u^{2} - 3 u) + (- 10 u + 30)

Factor out

u

from

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

u^{2} - 3 u

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

u^{2} = uu

= uu - 3 u

Factor out common term

u

= u (u - 3)

Factor out

- 10

from

- 10 u + 30 : - 10 (u - 3)

- 10 u + 30

Rewrite

30

10 \cdot 3

= - 10 u + 10 \cdot 3

Factor out common term

- 10

= - 10 (u - 3)

= u (u - 3) - 10 (u - 3)

Factor out common term

u - 3

= (u - 3) (u - 10)

= (u - 3) (u - 10)

Substitute back

u = v^{2}

= (v^{2} - 3) (v^{2} - 10)

Factor

v^{2} - 3 : (v + 3) (v - 3)

v^{2} - 3

Apply radical rule:

a = (a)^{2}

3 = (3)^{2}

= v^{2} - (3)^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

v^{2} - (3)^{2} = (v + 3) (v - 3)

= (v + 3) (v - 3)

= (v + 3) (v - 3) (v^{2} - 10)

Factor

v^{2} - 10 : (v + 10) (v - 10)

v^{2} - 10

Apply radical rule:

a = (a)^{2}

10 = (10)^{2}

= v^{2} - (10)^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

v^{2} - (10)^{2} = (v + 10) (v - 10)

= (v + 10) (v - 10)

= (v + 3) (v - 3) (v + 10) (v - 10)

(v + 3) (v - 3) (v + 10) (v - 10) \leq 0

Identify the intervals

Find the signs of the factors of

(v + 3) (v - 3) (v + 10) (v - 10)

Find the signs of

v + 3

v + 3 = 0 : v = - 3

v + 3 = 0

Move

3

to the right side

v + 3 = 0

Subtract

3

from both sides

v + 3 - 3 = 0 - 3

Simplify

v = - 3

v = - 3

v + 3 < 0 : v < - 3

v + 3 < 0

Move

3

to the right side

v + 3 < 0

Subtract

3

from both sides

v + 3 - 3 < 0 - 3

Simplify

v < - 3

v < - 3

v + 3 > 0 : v > - 3

v + 3 > 0

Move

3

to the right side

v + 3 > 0

Subtract

3

from both sides

v + 3 - 3 > 0 - 3

Simplify

v > - 3

v > - 3

Find the signs of

v - 3

v - 3 = 0 : v = 3

v - 3 = 0

Move

3

to the right side

v - 3 = 0

Add

3

to both sides

v - 3 + 3 = 0 + 3

Simplify

v = 3

v = 3

v - 3 < 0 : v < 3

v - 3 < 0

Move

3

to the right side

v - 3 < 0

Add

3

to both sides

v - 3 + 3 < 0 + 3

Simplify

v < 3

v < 3

v - 3 > 0 : v > 3

v - 3 > 0

Move

3

to the right side

v - 3 > 0

Add

3

to both sides

v - 3 + 3 > 0 + 3

Simplify

v > 3

v > 3

Find the signs of

v + 10

v + 10 = 0 : v = - 10

v + 10 = 0

Move

10

to the right side

v + 10 = 0

Subtract

10

from both sides

v + 10 - 10 = 0 - 10

Simplify

v = - 10

v = - 10

v + 10 < 0 : v < - 10

v + 10 < 0

Move

10

to the right side

v + 10 < 0

Subtract

10

from both sides

v + 10 - 10 < 0 - 10

Simplify

v < - 10

v < - 10

v + 10 > 0 : v > - 10

v + 10 > 0

Move

10

to the right side

v + 10 > 0

Subtract

10

from both sides

v + 10 - 10 > 0 - 10

Simplify

v > - 10

v > - 10

Find the signs of

v - 10

v - 10 = 0 : v = 10

v - 10 = 0

Move

10

to the right side

v - 10 = 0

Add

10

to both sides

v - 10 + 10 = 0 + 10

Simplify

v = 10

v = 10

v - 10 < 0 : v < 10

v - 10 < 0

Move

10

to the right side

v - 10 < 0

Add

10

to both sides

v - 10 + 10 < 0 + 10

Simplify

v < 10

v < 10

v - 10 > 0 : v > 10

v - 10 > 0

Move

10

to the right side

v - 10 > 0

Add

10

to both sides

v - 10 + 10 > 0 + 10

Simplify

v > 10

v > 10

Summarize in a table:

v + 3 v - 3 v + 10 v - 10 (v + 3) (v - 3) (v + 10) (v - 10) v < - 10 - - - - + v = - 10 - - 0 - 0 - 10 < v < - 3 - - + - - v = - 3 0 - + - 0 - 3 < v < 3 + - + - + v = 3 + 0 + - 0 3 < v < 10 + + + - - v = 10 + + + 00 v > 10 + + + + +

Identify the intervals that satisfy the required condition:

\leq 0

v = - 10 or - 10 < v < - 3 or v = - 3 or v = 3 or 3 < v < 10 or v = 10

Merge Overlapping Intervals

- 10 \leq v \leq - 3 or 3 \leq v < 10 or v = 10

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

v = - 10

- 10 < v < - 3

- 10 \leq v < - 3

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

- 10 \leq v < - 3

v = - 3

- 10 \leq v \leq - 3

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

- 10 \leq v \leq - 3

v = 3

- 10 \leq v \leq - 3 or v = 3

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

- 10 \leq v \leq - 3 or v = 3

3 < v < 10

- 10 \leq v \leq - 3 or 3 \leq v < 10

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

- 10 \leq v \leq - 3 or 3 \leq v < 10

v = 10

- 10 \leq v \leq - 3 or 3 \leq v \leq 10

- 10 \leq v \leq - 3 or 3 \leq v \leq 10

- 10 \leq v \leq - 3 or 3 \leq v \leq 10

- 10 \leq v \leq - 3 or 3 \leq v \leq 10

Substitute back

v = 3^{s i n^{2} (π x)}

- 10 \leq 3^{s i n^{2} (π x)} \leq - 3 or 3 \leq 3^{s i n^{2} (π x)} \leq 10

- 10 \leq 3^{s i n^{2} (π x)} \leq - 3 :

False for all

x \in R

- 10 \leq 3^{s i n^{2} (π x)} \leq - 3

a \leq u \leq b

then

a \leq u and u \leq b

- 10 \leq 3^{s i n^{2} (π x)} and 3^{s i n^{2} (π x)} \leq - 3

- 10 \leq 3^{s i n^{2} (π x)} :

True for all

x \in R

- 10 \leq 3^{s i n^{2} (π x)}

Switch sides

3^{s i n^{2} (π x)} \geq - 10

Apply exponent rules

3^{s i n^{2} (π x)} \geq - 10

a > 0, a^{f (x)}

is greater than 0

a = 3

True for all x \in R

True for all x

True for all x \in R

3^{s i n^{2} (π x)} \leq - 3 :

False for all

x \in R

3^{s i n^{2} (π x)} \leq - 3

Apply exponent rules

3^{s i n^{2} (π x)} \leq - 3

a > 0, a^{f (x)}

is greater than 0

a = 3

False for all x \in R

No Solution for x \in R

False for all x \in R

Combine the intervals

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

Merge Overlapping Intervals

True for all x \in R and False 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

False for all

x \in R

False for all x \in R

False for all x \in R

3 \leq 3^{s i n^{2} (π x)} \leq 10 :

False for all

x \in R

3 \leq 3^{s i n^{2} (π x)} \leq 10

a \leq u \leq b

then

a \leq u and u \leq b

3 \leq 3^{s i n^{2} (π x)} and 3^{s i n^{2} (π x)} \leq 10

3 \leq 3^{s i n^{2} (π x)} : arcsin (\frac{1}{2}) + 2 πn \leq π x \leq π - arcsin (\frac{1}{2}) + 2 πn

3 \leq 3^{s i n^{2} (π x)}

Apply exponent rules

3 \leq 3^{s i n^{2} (π x)}

a > 1,

then

a^{f (x)} \leq a^{g (x)}

is equivalent to

f (x) \leq g (x)

a = 3, f (x) = \frac{1}{2}, g (x) = sin^{2} (π x)

\frac{1}{2} \leq sin^{2} (π x)

\frac{1}{2} \leq sin^{2} (π x)

\frac{1}{2} \leq sin^{2} (π x) : arcsin (\frac{1}{2}) + 2 πn \leq π x \leq π - arcsin (\frac{1}{2}) + 2 πn

\frac{1}{2} \leq sin^{2} (π x)

Switch sides

sin^{2} (π x) \geq \frac{1}{2}

For

u^{n} \geq a

, if

n

is even then

u \leq - n a or u \geq n a

sin (π x) \leq - \frac{1}{2} or sin (π x) \geq \frac{1}{2}

sin (π x) \leq - \frac{1}{2} : - π - arcsin (- \frac{1}{2}) + 2 πn \leq π x \leq arcsin (- \frac{1}{2}) + 2 πn

sin (π x) \leq - \frac{1}{2}

For

sin (x) \leq a

, if

- 1 < a < 1

then

- π - arcsin (a) + 2 πn \leq x \leq arcsin (a) + 2 πn

- π - arcsin (- \frac{1}{2}) + 2 πn \leq π x \leq arcsin (- \frac{1}{2}) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

- π - arcsin (- \frac{1}{2}) + 2 πn \leq π x \leq arcsin (- \frac{1}{2}) + 2 πn

sin (π x) \geq \frac{1}{2} : arcsin (\frac{1}{2}) + 2 πn \leq π x \leq π - arcsin (\frac{1}{2}) + 2 πn

sin (π x) \geq \frac{1}{2}

For

sin (x) \geq a

, if

- 1 < a < 1

then

arcsin (a) + 2 πn \leq x \leq π - arcsin (a) + 2 πn

arcsin (\frac{1}{2}) + 2 πn \leq π x \leq π - arcsin (\frac{1}{2}) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

arcsin (\frac{1}{2}) + 2 πn \leq π x \leq π - arcsin (\frac{1}{2}) + 2 πn

Combine the intervals

- π - arcsin (- \frac{1}{2}) + 2 πn \leq π x \leq arcsin (- \frac{1}{2}) + 2 πn or arcsin (\frac{1}{2}) + 2 πn \leq π x \leq π - arcsin (\frac{1}{2}) + 2 πn

Merge Overlapping Intervals

arcsin (\frac{1}{2}) + 2 πn \leq π x \leq π - arcsin (\frac{1}{2}) + 2 πn

arcsin (\frac{1}{2}) + 2 πn \leq π x \leq π - arcsin (\frac{1}{2}) + 2 πn

3^{s i n^{2} (π x)} \leq 10 :

True for all

x \in R

3^{s i n^{2} (π x)} \leq 10

f (x) \leq g (x)

then

ln (f (x)) \leq ln (g (x))

ln (3^{s i n^{2} (π x)}) \leq ln (10)

Simplify

ln (3^{s i n^{2} (π x)}) : ln (3) sin^{2} (π x)

ln (3^{s i n^{2} (π x)})

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= ln (3) sin^{2} (π x)

Simplify

ln (10) : \frac{1}{2} ln (10)

ln (10)

Rewrite as

= ln (1 0^{\frac{1}{2}})

Apply log rule

lo g_{a} (x^{b}) = b \cdot lo g_{a} (x),

assuming

x \geq 0

= \frac{1}{2} ln (10)

ln (3) sin^{2} (π x) \leq \frac{1}{2} ln (10)

ln (3) sin^{2} (π x) \leq \frac{1}{2} ln (10) :

True for all

x

ln (3) sin^{2} (π x) \leq \frac{1}{2} ln (10)

Divide both sides by

ln (3)

ln (3) sin^{2} (π x) \leq \frac{1}{2} ln (10)

Divide both sides by

ln (3)

\frac{ln ( 3 ) sin ^{2} ( π x )}{ln ( 3 )} \leq \frac{\frac{1}{2} ln ( 10 )}{ln ( 3 )}

Simplify

\frac{ln ( 3 ) sin ^{2} ( π x )}{ln ( 3 )} \leq \frac{\frac{1}{2} ln ( 10 )}{ln ( 3 )}

Simplify

\frac{ln ( 3 ) sin ^{2} ( π x )}{ln ( 3 )} : sin^{2} (π x)

\frac{ln ( 3 ) sin ^{2} ( π x )}{ln ( 3 )}

Cancel the common factor:

ln (3)

= sin^{2} (π x)

Simplify

\frac{\frac{1}{2} ln ( 10 )}{ln ( 3 )} : \frac{ln ( 10 )}{2 ln ( 3 )}

\frac{\frac{1}{2} ln ( 10 )}{ln ( 3 )}

Multiply

\frac{1}{2} ln (10) : \frac{ln ( 10 )}{2}

\frac{1}{2} ln (10)

Multiply fractions:

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

= \frac{1 \cdot ln ( 10 )}{2}

Multiply:

1 \cdot ln (10) = ln (10)

= \frac{ln ( 10 )}{2}

= \frac{\frac{l n ( 10 )}{2}}{ln ( 3 )}

Apply the fraction rule:

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

= \frac{ln ( 10 )}{2 ln ( 3 )}

sin^{2} (π x) \leq \frac{ln ( 10 )}{2 ln ( 3 )}

sin^{2} (π x) \leq \frac{ln ( 10 )}{2 ln ( 3 )}

sin^{2} (π x) \leq \frac{ln ( 10 )}{2 ln ( 3 )}

For

u^{n} \leq a

, if

n

is even then

- n a \leq u \leq n a

- \frac{ln ( 10 )}{2 ln ( 3 )} \leq sin (π x) \leq \frac{ln ( 10 )}{2 ln ( 3 )}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{ln ( 10 )}{2 ln ( 3 )} \leq sin (π x) and sin (π x) \leq \frac{ln ( 10 )}{2 ln ( 3 )}

- \frac{ln ( 10 )}{2 ln ( 3 )} \leq sin (π x) :

True for all

x \in R

- \frac{ln ( 10 )}{2 ln ( 3 )} \leq sin (π x)

Switch sides

sin (π x) \geq - \frac{ln ( 10 )}{2 ln ( 3 )}

Range of

sin (π x) : - 1 \leq sin (π x) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (π x) \leq 1

- 1 \leq sin (π x) \leq 1

sin (π x) \geq - \frac{ln ( 10 )}{2 ln ( 3 )} and - 1 \leq sin (π x) \leq 1 : - 1 \leq sin (π x) \leq 1

Let

y = sin (π x)

Combine the intervals

y \geq - \frac{ln ( 10 )}{2 ln ( 3 )} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \geq - \frac{ln ( 10 )}{2 ln ( 3 )} and - 1 \leq y \leq 1

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

y \geq - \frac{ln ( 10 )}{2 ln ( 3 )}

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 (π x) \leq \frac{ln ( 10 )}{2 ln ( 3 )} :

True for all

x \in R

sin (π x) \leq \frac{ln ( 10 )}{2 ln ( 3 )}

Range of

sin (π x) : - 1 \leq sin (π x) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (π x) \leq 1

- 1 \leq sin (π x) \leq 1

sin (π x) \leq \frac{ln ( 10 )}{2 ln ( 3 )} and - 1 \leq sin (π x) \leq 1 : - 1 \leq sin (π x) \leq 1

Let

y = sin (π x)

Combine the intervals

y \leq \frac{ln ( 10 )}{2 ln ( 3 )} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \leq \frac{ln ( 10 )}{2 ln ( 3 )} and - 1 \leq y \leq 1

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

y \leq \frac{ln ( 10 )}{2 ln ( 3 )}

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

Combine the intervals

arcsin (\frac{1}{2}) + 2 πn \leq π x \leq π - arcsin (\frac{1}{2}) + 2 πn and True for all x \in R

Merge Overlapping Intervals

False 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

False for all

x \in R

and

True for all

x \in R

False for all x \in R

False for all x \in R

Combine the intervals

False for all x \in R or False for all x \in R

Merge Overlapping Intervals

False for all x \in R or False for all x \in R

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

False for all

x \in R

False for all

x \in R

False for all x \in R

No Solution for x \in R

False for all x \in R

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