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Popular Trigonometry >

sin^2(3x)-cos^2(3x)<= (sqrt(3))/2

Solution

sin^{2} (3 x) - cos^{2} (3 x) \leq \frac{3}{2}

Solution

\frac{2 π}{3} n \leq x \leq \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n or \frac{π - arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n \leq x \leq \frac{π + arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n or \frac{- arcsin ( \frac{2 + 3}{2} ) + 2 π}{3} + \frac{2 π}{3} n \leq x < \frac{2 π}{3} + \frac{2 π}{3} n

Interval Notation

\frac{2 π}{3} n, \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n \cup \frac{π - arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n, \frac{π + arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n \cup \frac{- arcsin ( \frac{2 + 3}{2} ) + 2 π}{3} + \frac{2 π}{3} n, \frac{2 π}{3} + \frac{2 π}{3} n

Decimal

\frac{2 π}{3} n \leq x \leq 0.43633 \dots + \frac{2 π}{3} n or 0.61086 \dots + \frac{2 π}{3} n \leq x \leq 1.48352 \dots + \frac{2 π}{3} n or 1.65806 \dots + \frac{2 π}{3} n \leq x < 2.09439 \dots + \frac{2 π}{3} n

Solution steps

sin^{2} (3 x) - cos^{2} (3 x) \leq \frac{3}{2}

Use the following identity:

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

Therefore

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

sin^{2} (3 x) - (1 - sin^{2} (3 x)) \leq \frac{3}{2}

Simplify

2 sin^{2} (3 x) - 1 \leq \frac{3}{2}

Rewrite in standard form

2 sin^{2} (3 x) - 1 \leq \frac{3}{2}

Subtract

\frac{3}{2}

from both sides

2 sin^{2} (3 x) - 1 - \frac{3}{2} \leq \frac{3}{2} - \frac{3}{2}

Simplify

2 sin^{2} (3 x) - 1 - \frac{3}{2} \leq 0

Multiply both sides by

2

2 sin^{2} (3 x) \cdot 2 - 1 \cdot 2 - \frac{3}{2} \cdot 2 \leq 0 \cdot 2

4 sin^{2} (3 x) - 2 - 3 \leq 0

4 sin^{2} (3 x) - 2 - 3 \leq 0

Identify the intervals

Find the signs of the factors of

4 sin^{2} (3 x) - 2 - 3

Find the signs of

4 sin^{2} (3 x) - 2 - 3

4 sin^{2} (3 x) - 2 - 3 = 0 : sin (3 x) = - \frac{1}{2} + \frac{3}{4} or sin (3 x) = \frac{1}{2} + \frac{3}{4}

4 sin^{2} (3 x) - 2 - 3 = 0

Move

2

to the right side

4 sin^{2} (3 x) - 2 - 3 = 0

Add

2

to both sides

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

Simplify

4 sin^{2} (3 x) - 3 = 2

4 sin^{2} (3 x) - 3 = 2

Move

3

to the right side

4 sin^{2} (3 x) - 3 = 2

Add

3

to both sides

4 sin^{2} (3 x) - 3 + 3 = 2 + 3

Simplify

4 sin^{2} (3 x) = 2 + 3

4 sin^{2} (3 x) = 2 + 3

Divide both sides by

4

4 sin^{2} (3 x) = 2 + 3

Divide both sides by

4

\frac{4 sin ^{2} ( 3 x )}{4} = \frac{2}{4} + \frac{3}{4}

Simplify

sin^{2} (3 x) = \frac{1}{2} + \frac{3}{4}

sin^{2} (3 x) = \frac{1}{2} + \frac{3}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

sin (3 x) = \frac{1}{2} + \frac{3}{4}, sin (3 x) = - \frac{1}{2} + \frac{3}{4}

4 sin^{2} (3 x) - 2 - 3 < 0 : - \frac{1}{2} + \frac{3}{4} < sin (3 x) < \frac{1}{2} + \frac{3}{4}

4 sin^{2} (3 x) - 2 - 3 < 0

Identify the intervals

Find the signs of the factors of

4 sin^{2} (3 x) - 2 - 3

Find the signs of

4 sin^{2} (3 x) - 2 - 3

4 sin^{2} (3 x) - 2 - 3 = 0 : sin (3 x) = - \frac{1}{2} + \frac{3}{4} or sin (3 x) = \frac{1}{2} + \frac{3}{4}

4 sin^{2} (3 x) - 2 - 3 = 0

Move

2

to the right side

4 sin^{2} (3 x) - 2 - 3 = 0

Add

2

to both sides

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

Simplify

4 sin^{2} (3 x) - 3 = 2

4 sin^{2} (3 x) - 3 = 2

Move

3

to the right side

4 sin^{2} (3 x) - 3 = 2

Add

3

to both sides

4 sin^{2} (3 x) - 3 + 3 = 2 + 3

Simplify

4 sin^{2} (3 x) = 2 + 3

4 sin^{2} (3 x) = 2 + 3

Divide both sides by

4

4 sin^{2} (3 x) = 2 + 3

Divide both sides by

4

\frac{4 sin ^{2} ( 3 x )}{4} = \frac{2}{4} + \frac{3}{4}

Simplify

sin^{2} (3 x) = \frac{1}{2} + \frac{3}{4}

sin^{2} (3 x) = \frac{1}{2} + \frac{3}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

sin (3 x) = \frac{1}{2} + \frac{3}{4}, sin (3 x) = - \frac{1}{2} + \frac{3}{4}

Summarize in a table:

4 sin^{2} (3 x) - 2 - 3 4 sin^{2} (3 x) - 2 - 3 sin (3 x) < - \frac{1}{2} + \frac{3}{4} + + sin (3 x) = - \frac{1}{2} + \frac{3}{4} 00 - \frac{1}{2} + \frac{3}{4} < sin (3 x) < \frac{1}{2} + \frac{3}{4} - - sin (3 x) = \frac{1}{2} + \frac{3}{4} 00 sin (3 x) > \frac{1}{2} + \frac{3}{4} + +

Identify the intervals that satisfy the required condition:

< 0

- \frac{1}{2} + \frac{3}{4} < sin (3 x) < \frac{1}{2} + \frac{3}{4}

- \frac{1}{2} + \frac{3}{4} < sin (3 x) < \frac{1}{2} + \frac{3}{4}

4 sin^{2} (3 x) - 2 - 3 > 0 : sin (3 x) < - \frac{1}{2} + \frac{3}{4} or sin (3 x) > \frac{1}{2} + \frac{3}{4}

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

Identify the intervals

Find the signs of the factors of

4 sin^{2} (3 x) - 2 - 3

Find the signs of

4 sin^{2} (3 x) - 2 - 3

4 sin^{2} (3 x) - 2 - 3 = 0 : sin (3 x) = - \frac{1}{2} + \frac{3}{4} or sin (3 x) = \frac{1}{2} + \frac{3}{4}

4 sin^{2} (3 x) - 2 - 3 = 0

Move

2

to the right side

4 sin^{2} (3 x) - 2 - 3 = 0

Add

2

to both sides

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

Simplify

4 sin^{2} (3 x) - 3 = 2

4 sin^{2} (3 x) - 3 = 2

Move

3

to the right side

4 sin^{2} (3 x) - 3 = 2

Add

3

to both sides

4 sin^{2} (3 x) - 3 + 3 = 2 + 3

Simplify

4 sin^{2} (3 x) = 2 + 3

4 sin^{2} (3 x) = 2 + 3

Divide both sides by

4

4 sin^{2} (3 x) = 2 + 3

Divide both sides by

4

\frac{4 sin ^{2} ( 3 x )}{4} = \frac{2}{4} + \frac{3}{4}

Simplify

sin^{2} (3 x) = \frac{1}{2} + \frac{3}{4}

sin^{2} (3 x) = \frac{1}{2} + \frac{3}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

sin (3 x) = \frac{1}{2} + \frac{3}{4}, sin (3 x) = - \frac{1}{2} + \frac{3}{4}

Summarize in a table:

4 sin^{2} (3 x) - 2 - 3 4 sin^{2} (3 x) - 2 - 3 sin (3 x) < - \frac{1}{2} + \frac{3}{4} + + sin (3 x) = - \frac{1}{2} + \frac{3}{4} 00 - \frac{1}{2} + \frac{3}{4} < sin (3 x) < \frac{1}{2} + \frac{3}{4} - - sin (3 x) = \frac{1}{2} + \frac{3}{4} 00 sin (3 x) > \frac{1}{2} + \frac{3}{4} + +

Identify the intervals that satisfy the required condition:

> 0

sin (3 x) < - \frac{1}{2} + \frac{3}{4} or sin (3 x) > \frac{1}{2} + \frac{3}{4}

sin (3 x) < - \frac{1}{2} + \frac{3}{4} or sin (3 x) > \frac{1}{2} + \frac{3}{4}

Summarize in a table:

4 sin^{2} (3 x) - 2 - 3 4 sin^{2} (3 x) - 2 - 3 sin (3 x) < - \frac{1}{2} + \frac{3}{4} + + sin (3 x) = - \frac{1}{2} + \frac{3}{4} 00 - \frac{1}{2} + \frac{3}{4} < sin (3 x) < \frac{1}{2} + \frac{3}{4} - - sin (3 x) = \frac{1}{2} + \frac{3}{4} 00 sin (3 x) > \frac{1}{2} + \frac{3}{4} + +

Identify the intervals that satisfy the required condition:

\leq 0

sin (3 x) = - \frac{1}{2} + \frac{3}{4} or - \frac{1}{2} + \frac{3}{4} < sin (3 x) < \frac{1}{2} + \frac{3}{4} or sin (3 x) = \frac{1}{2} + \frac{3}{4}

Merge Overlapping Intervals

- \frac{1}{2} + \frac{3}{4} \leq sin (3 x) < \frac{1}{2} + \frac{3}{4} or sin (3 x) = \frac{1}{2} + \frac{3}{4}

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

sin (3 x) = - \frac{1}{2} + \frac{3}{4}

- \frac{1}{2} + \frac{3}{4} < sin (3 x) < \frac{1}{2} + \frac{3}{4}

- \frac{1}{2} + \frac{3}{4} \leq sin (3 x) < \frac{1}{2} + \frac{3}{4}

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

- \frac{1}{2} + \frac{3}{4} \leq sin (3 x) < \frac{1}{2} + \frac{3}{4}

sin (3 x) = \frac{1}{2} + \frac{3}{4}

- \frac{1}{2} + \frac{3}{4} \leq sin (3 x) \leq \frac{1}{2} + \frac{3}{4}

- \frac{1}{2} + \frac{3}{4} \leq sin (3 x) \leq \frac{1}{2} + \frac{3}{4}

- \frac{1}{2} + \frac{3}{4} \leq sin (3 x) \leq \frac{1}{2} + \frac{3}{4}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{1}{2} + \frac{3}{4} \leq sin (3 x) and sin (3 x) \leq \frac{1}{2} + \frac{3}{4}

- \frac{1}{2} + \frac{3}{4} \leq sin (3 x) : - \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n \leq x \leq \frac{π + arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

- \frac{1}{2} + \frac{3}{4} \leq sin (3 x)

Switch sides

sin (3 x) \geq - \frac{1}{2} + \frac{3}{4}

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} + \frac{3}{4} + 2 πn \leq 3 x \leq π - arcsin - \frac{1}{2} + \frac{3}{4} + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

arcsin - \frac{1}{2} + \frac{3}{4} + 2 πn \leq 3 x and 3 x \leq π - arcsin - \frac{1}{2} + \frac{3}{4} + 2 πn

arcsin - \frac{1}{2} + \frac{3}{4} + 2 πn \leq 3 x : x \geq - \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

arcsin - \frac{1}{2} + \frac{3}{4} + 2 πn \leq 3 x

Switch sides

3 x \geq arcsin - \frac{1}{2} + \frac{3}{4} + 2 πn

Simplify

arcsin - \frac{1}{2} + \frac{3}{4} + 2 πn : - arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

arcsin - \frac{1}{2} + \frac{3}{4} + 2 πn

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin - \frac{1}{2} + \frac{3}{4} = - arcsin \frac{1}{2} + \frac{3}{4}

= - arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

3 x \geq - arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

Divide both sides by

3

3 x \geq - arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

Divide both sides by

3

\frac{3 x}{3} \geq - \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} + \frac{2 πn}{3}

Simplify

x \geq - \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} + \frac{2 πn}{3}

Simplify

- \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} : - \frac{arcsin ( \frac{2 + 3}{2} )}{3}

- \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3}

\frac{1}{2} + \frac{3}{4} = \frac{2 + 3}{2}

\frac{1}{2} + \frac{3}{4}

Join

\frac{1}{2} + \frac{3}{4} : \frac{2 + 3}{4}

\frac{1}{2} + \frac{3}{4}

Least Common Multiplier of

2, 4 : 4

2, 4

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

2

4

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{1}{2} :

multiply the denominator and numerator by

2

\frac{1}{2} = \frac{1 \cdot 2}{2 \cdot 2} = \frac{2}{4}

= \frac{2}{4} + \frac{3}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{2 + 3}{4}

= \frac{2 + 3}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{2 + 3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2 + 3}{2}

= - \frac{arcsin ( \frac{2 + 3}{2} )}{3}

x \geq - \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

x \geq - \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

3 x \leq π - arcsin - \frac{1}{2} + \frac{3}{4} + 2 πn : x \leq \frac{π + arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

3 x \leq π - arcsin - \frac{1}{2} + \frac{3}{4} + 2 πn

Simplify

π - arcsin - \frac{1}{2} + \frac{3}{4} + 2 πn : π + arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

π - arcsin - \frac{1}{2} + \frac{3}{4} + 2 πn

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin - \frac{1}{2} + \frac{3}{4} = - arcsin \frac{1}{2} + \frac{3}{4}

= π - - arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

Apply rule

- (- a) = a

= π + arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

3 x \leq π + arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

Divide both sides by

3

3 x \leq π + arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

Divide both sides by

3

\frac{3 x}{3} \leq \frac{π}{3} + \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} + \frac{2 πn}{3}

Simplify

x \leq \frac{π}{3} + \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} + \frac{2 πn}{3}

Simplify

\frac{π}{3} + \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} : \frac{π + arcsin ( \frac{2 + 3}{2} )}{3}

\frac{π}{3} + \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3}

Combine the fractions

\frac{π}{3} + \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} : \frac{π + arcsin ( \frac{2 + 3}{2} )}{3}

Apply rule

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

= \frac{π + arcsin ( \frac{1}{2} + \frac{3}{4} )}{3}

\frac{1}{2} + \frac{3}{4} = \frac{2 + 3}{2}

\frac{1}{2} + \frac{3}{4}

Join

\frac{1}{2} + \frac{3}{4} : \frac{2 + 3}{4}

\frac{1}{2} + \frac{3}{4}

Least Common Multiplier of

2, 4 : 4

2, 4

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

2

4

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{1}{2} :

multiply the denominator and numerator by

2

\frac{1}{2} = \frac{1 \cdot 2}{2 \cdot 2} = \frac{2}{4}

= \frac{2}{4} + \frac{3}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{2 + 3}{4}

= \frac{2 + 3}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{2 + 3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2 + 3}{2}

= \frac{2 + 3}{2}

= \frac{π + arcsin ( \frac{2 + 3}{2} )}{3}

x \leq \frac{π + arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

x \leq \frac{π + arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

x \leq \frac{π + arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

Combine the intervals

x \geq - \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n and x \leq \frac{π + arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

Merge Overlapping Intervals

- \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n \leq x \leq \frac{π + arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

sin (3 x) \leq \frac{1}{2} + \frac{3}{4} : \frac{- π - arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n \leq x \leq \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

sin (3 x) \leq \frac{1}{2} + \frac{3}{4}

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} + \frac{3}{4} + 2 πn \leq 3 x \leq arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

- π - arcsin \frac{1}{2} + \frac{3}{4} + 2 πn \leq 3 x and 3 x \leq arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

- π - arcsin \frac{1}{2} + \frac{3}{4} + 2 πn \leq 3 x : x \geq \frac{- π - arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

- π - arcsin \frac{1}{2} + \frac{3}{4} + 2 πn \leq 3 x

Switch sides

3 x \geq - π - arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

Divide both sides by

3

3 x \geq - π - arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

Divide both sides by

3

\frac{3 x}{3} \geq - \frac{π}{3} - \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} + \frac{2 πn}{3}

Simplify

x \geq - \frac{π}{3} - \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} + \frac{2 πn}{3}

Simplify

- \frac{π}{3} - \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} : \frac{- π - arcsin ( \frac{2 + 3}{2} )}{3}

- \frac{π}{3} - \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3}

Combine the fractions

- \frac{π}{3} - \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} : \frac{- π - arcsin ( \frac{2 + 3}{2} )}{3}

Apply rule

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

= \frac{- π - arcsin ( \frac{1}{2} + \frac{3}{4} )}{3}

\frac{1}{2} + \frac{3}{4} = \frac{2 + 3}{2}

\frac{1}{2} + \frac{3}{4}

Join

\frac{1}{2} + \frac{3}{4} : \frac{2 + 3}{4}

\frac{1}{2} + \frac{3}{4}

Least Common Multiplier of

2, 4 : 4

2, 4

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

2

4

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{1}{2} :

multiply the denominator and numerator by

2

\frac{1}{2} = \frac{1 \cdot 2}{2 \cdot 2} = \frac{2}{4}

= \frac{2}{4} + \frac{3}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{2 + 3}{4}

= \frac{2 + 3}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{2 + 3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2 + 3}{2}

= \frac{2 + 3}{2}

= \frac{- π - arcsin ( \frac{2 + 3}{2} )}{3}

x \geq \frac{- π - arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

x \geq \frac{- π - arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

x \geq \frac{- π - arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

3 x \leq arcsin \frac{1}{2} + \frac{3}{4} + 2 πn : x \leq \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

3 x \leq arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

Divide both sides by

3

3 x \leq arcsin \frac{1}{2} + \frac{3}{4} + 2 πn

Divide both sides by

3

\frac{3 x}{3} \leq \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} + \frac{2 πn}{3}

Simplify

x \leq \frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} + \frac{2 πn}{3}

Simplify

\frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3} : \frac{arcsin ( \frac{2 + 3}{2} )}{3}

\frac{arcsin ( \frac{1}{2} + \frac{3}{4} )}{3}

\frac{1}{2} + \frac{3}{4} = \frac{2 + 3}{2}

\frac{1}{2} + \frac{3}{4}

Join

\frac{1}{2} + \frac{3}{4} : \frac{2 + 3}{4}

\frac{1}{2} + \frac{3}{4}

Least Common Multiplier of

2, 4 : 4

2, 4

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

2

4

= 2 \cdot 2

Multiply the numbers:

2 \cdot 2 = 4

= 4

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

4

For

\frac{1}{2} :

multiply the denominator and numerator by

2

\frac{1}{2} = \frac{1 \cdot 2}{2 \cdot 2} = \frac{2}{4}

= \frac{2}{4} + \frac{3}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{2 + 3}{4}

= \frac{2 + 3}{4}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{2 + 3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

n a^{n} = a

2^{2} = 2

= 2

= \frac{2 + 3}{2}

= \frac{arcsin ( \frac{2 + 3}{2} )}{3}

x \leq \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

x \leq \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

Combine the intervals

x \geq \frac{- π - arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n and x \leq \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

Merge Overlapping Intervals

\frac{- π - arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n \leq x \leq \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

Combine the intervals

- \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n \leq x \leq \frac{π + arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n and \frac{- π - arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n \leq x \leq \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n

Merge Overlapping Intervals

\frac{2 π}{3} n \leq x \leq \frac{arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n or \frac{π - arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n \leq x \leq \frac{π + arcsin ( \frac{2 + 3}{2} )}{3} + \frac{2 π}{3} n or \frac{- arcsin ( \frac{2 + 3}{2} ) + 2 π}{3} + \frac{2 π}{3} n \leq x < \frac{2 π}{3} + \frac{2 π}{3} n

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