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Popular Trigonometry >

sin^2(2x)<= 1/2

Solution

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

Solution

πn \leq x \leq \frac{π}{8} + πn or \frac{3 π}{8} + πn \leq x \leq \frac{5 π}{8} + πn or \frac{7 π}{8} + πn \leq x < π + πn

Interval Notation

[πn, \frac{π}{8} + πn] \cup [\frac{3 π}{8} + πn, \frac{5 π}{8} + πn] \cup [\frac{7 π}{8} + πn, π + πn)

Decimal

πn \leq x \leq 0.39269 \dots + πn or 1.17809 \dots + πn \leq x \leq 1.96349 \dots + πn or 2.74889 \dots + πn \leq x < 3.14159 \dots + πn

Solution steps

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

For

u^{n} \leq a

, if

n

is even then

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

a \leq u \leq b

then

a \leq u and u \leq b

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

- \frac{1}{2} \leq sin (2 x) : - \frac{π}{8} + πn \leq x \leq \frac{5 π}{8} + πn

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

Switch sides

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

arcsin (- \frac{1}{2}) + 2 πn \leq 2 x : x \geq - \frac{π}{8} + πn

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

Switch sides

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

Simplify

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

arcsin (- \frac{1}{2}) + 2 πn

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

arcsin (- \frac{1}{2})

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{4}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{4}

= - \frac{π}{4}

= - \frac{π}{4} + 2 πn

2 x \geq - \frac{π}{4} + 2 πn

Divide both sides by

2

2 x \geq - \frac{π}{4} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

- \frac{\frac{π}{4}}{2} + \frac{2 πn}{2} : - \frac{π}{8} + πn

- \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= - \frac{π}{8} + πn

x \geq - \frac{π}{8} + πn

x \geq - \frac{π}{8} + πn

x \geq - \frac{π}{8} + πn

2 x \leq π - arcsin (- \frac{1}{2}) + 2 πn : x \leq \frac{5 π}{8} + πn

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

Simplify

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

π - arcsin (- \frac{1}{2}) + 2 πn

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

arcsin (- \frac{1}{2})

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{2}) = - arcsin (\frac{1}{2})

= - arcsin (\frac{1}{2})

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{4}

arcsin (\frac{1}{2})

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{4}

= - \frac{π}{4}

= π - (- \frac{π}{4}) + 2 πn

Apply rule

- (- a) = a

= π + \frac{π}{4} + 2 πn

2 x \leq π + \frac{π}{4} + 2 πn

Divide both sides by

2

2 x \leq π + \frac{π}{4} + 2 πn

Divide both sides by

2

\frac{2 x}{2} \leq \frac{π}{2} + \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} \leq \frac{π}{2} + \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{π}{2} + \frac{\frac{π}{4}}{2} + \frac{2 πn}{2} : \frac{π}{2} + \frac{π}{8} + πn

\frac{π}{2} + \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{2} + \frac{π}{8} + πn

x \leq \frac{π}{2} + \frac{π}{8} + πn

x \leq \frac{π}{2} + \frac{π}{8} + πn

Simplify

\frac{π}{2} + \frac{π}{8} : \frac{5 π}{8}

\frac{π}{2} + \frac{π}{8}

Least Common Multiplier of

2, 8 : 8

2, 8

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

8 : 2 \cdot 2 \cdot 2

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

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

2

8

= 2 \cdot 2 \cdot 2

Multiply the numbers:

2 \cdot 2 \cdot 2 = 8

= 8

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

8

For

\frac{π}{2} :

multiply the denominator and numerator by

4

\frac{π}{2} = \frac{π 4}{2 \cdot 4} = \frac{π 4}{8}

= \frac{π 4}{8} + \frac{π}{8}

Since the denominators are equal, combine the fractions:

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

= \frac{π 4 + π}{8}

Add similar elements:

4 π + π = 5 π

= \frac{5 π}{8}

x \leq \frac{5 π}{8} + πn

x \leq \frac{5 π}{8} + πn

Combine the intervals

x \geq - \frac{π}{8} + πn and x \leq \frac{5 π}{8} + πn

Merge Overlapping Intervals

- \frac{π}{8} + πn \leq x \leq \frac{5 π}{8} + πn

sin (2 x) \leq \frac{1}{2} : - \frac{5 π}{8} + πn \leq x \leq \frac{π}{8} + πn

sin (2 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 2 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 2 x and 2 x \leq arcsin (\frac{1}{2}) + 2 πn

- π - arcsin (\frac{1}{2}) + 2 πn \leq 2 x : x \geq - \frac{5 π}{8} + πn

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

Switch sides

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

Simplify

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

- π - arcsin (\frac{1}{2}) + 2 πn

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{4}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= - π - \frac{π}{4} + 2 πn

2 x \geq - π - \frac{π}{4} + 2 πn

Divide both sides by

2

2 x \geq - π - \frac{π}{4} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

- \frac{π}{2} - \frac{\frac{π}{4}}{2} + \frac{2 πn}{2} : - \frac{π}{2} - \frac{π}{8} + πn

- \frac{π}{2} - \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= - \frac{π}{2} - \frac{π}{8} + πn

x \geq - \frac{π}{2} - \frac{π}{8} + πn

x \geq - \frac{π}{2} - \frac{π}{8} + πn

Simplify

- \frac{π}{2} - \frac{π}{8} : - \frac{5 π}{8}

- \frac{π}{2} - \frac{π}{8}

Least Common Multiplier of

2, 8 : 8

2, 8

Least Common Multiplier (LCM)

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

8 : 2 \cdot 2 \cdot 2

8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2

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

2

8

= 2 \cdot 2 \cdot 2

Multiply the numbers:

2 \cdot 2 \cdot 2 = 8

= 8

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

8

For

\frac{π}{2} :

multiply the denominator and numerator by

4

\frac{π}{2} = \frac{π 4}{2 \cdot 4} = \frac{π 4}{8}

= - \frac{π 4}{8} - \frac{π}{8}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 4 - π}{8}

Add similar elements:

- 4 π - π = - 5 π

= \frac{- 5 π}{8}

Apply the fraction rule:

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

= - \frac{5 π}{8}

x \geq - \frac{5 π}{8} + πn

x \geq - \frac{5 π}{8} + πn

2 x \leq arcsin (\frac{1}{2}) + 2 πn : x \leq \frac{π}{8} + πn

2 x \leq arcsin (\frac{1}{2}) + 2 πn

Simplify

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

arcsin (\frac{1}{2}) + 2 πn

Use the following trivial identity:

arcsin (\frac{1}{2}) = \frac{π}{4}

x 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arcsin (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} arcsin (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ}

= \frac{π}{4} + 2 πn

2 x \leq \frac{π}{4} + 2 πn

Divide both sides by

2

2 x \leq \frac{π}{4} + 2 πn

Divide both sides by

2

\frac{2 x}{2} \leq \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} \leq \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{4}}{2} + \frac{2 πn}{2} : \frac{π}{8} + πn

\frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{8} + πn

x \leq \frac{π}{8} + πn

x \leq \frac{π}{8} + πn

x \leq \frac{π}{8} + πn

Combine the intervals

x \geq - \frac{5 π}{8} + πn and x \leq \frac{π}{8} + πn

Merge Overlapping Intervals

- \frac{5 π}{8} + πn \leq x \leq \frac{π}{8} + πn

Combine the intervals

- \frac{π}{8} + πn \leq x \leq \frac{5 π}{8} + πn and - \frac{5 π}{8} + πn \leq x \leq \frac{π}{8} + πn

Merge Overlapping Intervals

πn \leq x \leq \frac{π}{8} + πn or \frac{3 π}{8} + πn \leq x \leq \frac{5 π}{8} + πn or \frac{7 π}{8} + πn \leq x < π + πn

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