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Popular >

2cos^2(x)< 3/2

Solution

2 cos^{2} (x) < \frac{3}{2}

Solution

\frac{π}{6} + 2 πn < x < \frac{5 π}{6} + 2 πn or \frac{7 π}{6} + 2 πn < x < \frac{11 π}{6} + 2 πn

Interval Notation

(\frac{π}{6} + 2 πn, \frac{5 π}{6} + 2 πn) \cup (\frac{7 π}{6} + 2 πn, \frac{11 π}{6} + 2 πn)

Decimal Notation

0.52359 \dots + 2 πn < x < 2.61799 \dots + 2 πn or 3.66519 \dots + 2 πn < x < 5.75958 \dots + 2 πn

Solution steps

2 cos^{2} (x) < \frac{3}{2}

Divide both sides by

2

2 cos^{2} (x) < \frac{3}{2}

Divide both sides by

2

\frac{2 cos ^{2} ( x )}{2} < \frac{\frac{3}{2}}{2}

Simplify

\frac{2 cos ^{2} ( x )}{2} < \frac{\frac{3}{2}}{2}

Simplify

\frac{2 cos ^{2} ( x )}{2} : cos^{2} (x)

\frac{2 cos ^{2} ( x )}{2}

Divide the numbers:

\frac{2}{2} = 1

= cos^{2} (x)

Simplify

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

\frac{\frac{3}{2}}{2}

Apply the fraction rule:

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

= \frac{3}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{3}{4}

cos^{2} (x) < \frac{3}{4}

cos^{2} (x) < \frac{3}{4}

cos^{2} (x) < \frac{3}{4}

For

u^{n} < a

, if

n

is even then

- \frac{3}{4} < cos (x) < \frac{3}{4}

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

\frac{3}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{3}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{3}{2}

- \frac{3}{2} < cos (x) < \frac{3}{2}

a < u < b

then

a < u and u < b

- \frac{3}{2} < cos (x) and cos (x) < \frac{3}{2}

- \frac{3}{2} < cos (x) : - \frac{5 π}{6} + 2 πn < x < \frac{5 π}{6} + 2 πn

- \frac{3}{2} < cos (x)

Switch sides

cos (x) > - \frac{3}{2}

For

cos (x) > a

, if

- 1 \leq a < 1

then

- arccos (a) + 2 πn < x < arccos (a) + 2 πn

- arccos (- \frac{3}{2}) + 2 πn < x < arccos (- \frac{3}{2}) + 2 πn

Simplify

- arccos (- \frac{3}{2}) : - \frac{5 π}{6}

- arccos (- \frac{3}{2})

Use the following trivial identity:

arccos (- \frac{3}{2}) = \frac{5 π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{5 π}{6}

Simplify

arccos (- \frac{3}{2}) : \frac{5 π}{6}

arccos (- \frac{3}{2})

Use the following trivial identity:

arccos (- \frac{3}{2}) = \frac{5 π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{5 π}{6}

- \frac{5 π}{6} + 2 πn < x < \frac{5 π}{6} + 2 πn

cos (x) < \frac{3}{2} : \frac{π}{6} + 2 πn < x < \frac{11 π}{6} + 2 πn

cos (x) < \frac{3}{2}

For

cos (x) < a

, if

- 1 < a \leq 1

then

arccos (a) + 2 πn < x < 2 π - arccos (a) + 2 πn

arccos (\frac{3}{2}) + 2 πn < x < 2 π - arccos (\frac{3}{2}) + 2 πn

Simplify

arccos (\frac{3}{2}) : \frac{π}{6}

arccos (\frac{3}{2})

Use the following trivial identity:

arccos (\frac{3}{2}) = \frac{π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{6}

Simplify

2 π - arccos (\frac{3}{2}) : \frac{11 π}{6}

2 π - arccos (\frac{3}{2})

Use the following trivial identity:

arccos (\frac{3}{2}) = \frac{π}{6}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{6}

Simplify

2 π - \frac{π}{6}

Convert element to fraction:

2 π = \frac{2 π 6}{6}

= \frac{2 π 6}{6} - \frac{π}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 6 - π}{6}

2 π 6 - π = 11 π

2 π 6 - π

Multiply the numbers:

2 \cdot 6 = 12

= 12 π - π

Add similar elements:

12 π - π = 11 π

= 11 π

= \frac{11 π}{6}

= \frac{11 π}{6}

\frac{π}{6} + 2 πn < x < \frac{11 π}{6} + 2 πn

Combine the intervals

- \frac{5 π}{6} + 2 πn < x < \frac{5 π}{6} + 2 πn and \frac{π}{6} + 2 πn < x < \frac{11 π}{6} + 2 πn

Merge Overlapping Intervals

\frac{π}{6} + 2 πn < x < \frac{5 π}{6} + 2 πn or \frac{7 π}{6} + 2 πn < x < \frac{11 π}{6} + 2 πn

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