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Popular >

2cos^2(x)-1>0

Solution

2 cos^{2} (x) - 1 > 0

Solution

- \frac{π}{4} + 2 πn < x < \frac{π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < \frac{5 π}{4} + 2 πn

Interval Notation

(- \frac{π}{4} + 2 πn, \frac{π}{4} + 2 πn) \cup (\frac{3 π}{4} + 2 πn, \frac{5 π}{4} + 2 πn)

Decimal Notation

- 0.78539 \dots + 2 πn < x < 0.78539 \dots + 2 πn or 2.35619 \dots + 2 πn < x < 3.92699 \dots + 2 πn

Solution steps

2 cos^{2} (x) - 1 > 0

Move

1

to the right side

2 cos^{2} (x) - 1 > 0

Add

1

to both sides

2 cos^{2} (x) - 1 + 1 > 0 + 1

Simplify

2 cos^{2} (x) > 1

2 cos^{2} (x) > 1

Divide both sides by

2

2 cos^{2} (x) > 1

Divide both sides by

2

\frac{2 cos ^{2} ( x )}{2} > \frac{1}{2}

Simplify

cos^{2} (x) > \frac{1}{2}

cos^{2} (x) > \frac{1}{2}

For

u^{n} > a

, if

n

is even then

cos (x) < - \frac{1}{2} or cos (x) > \frac{1}{2}

cos (x) < - \frac{1}{2} : \frac{3 π}{4} + 2 πn < x < \frac{5 π}{4} + 2 πn

cos (x) < - \frac{1}{2}

For

cos (x) < a

, if

- 1 < a \leq 1

then

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

arccos (- \frac{1}{2}) + 2 πn < x < 2 π - arccos (- \frac{1}{2}) + 2 πn

Simplify

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

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

Use the following trivial identity:

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

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{3 π}{4}

Simplify

2 π - arccos (- \frac{1}{2}) : \frac{5 π}{4}

2 π - arccos (- \frac{1}{2})

Use the following trivial identity:

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

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{3 π}{4}

Simplify

2 π - \frac{3 π}{4}

Convert element to fraction:

2 π = \frac{2 π 4}{4}

= \frac{2 π 4}{4} - \frac{3 π}{4}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 4 - 3 π}{4}

2 π 4 - 3 π = 5 π

2 π 4 - 3 π

Multiply the numbers:

2 \cdot 4 = 8

= 8 π - 3 π

Add similar elements:

8 π - 3 π = 5 π

= 5 π

= \frac{5 π}{4}

= \frac{5 π}{4}

\frac{3 π}{4} + 2 πn < x < \frac{5 π}{4} + 2 πn

cos (x) > \frac{1}{2} : - \frac{π}{4} + 2 πn < x < \frac{π}{4} + 2 πn

cos (x) > \frac{1}{2}

For

cos (x) > a

, if

- 1 \leq a < 1

then

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

- arccos (\frac{1}{2}) + 2 πn < x < arccos (\frac{1}{2}) + 2 πn

Simplify

- arccos (\frac{1}{2}) : - \frac{π}{4}

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

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{4}

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}{4}

Simplify

arccos (\frac{1}{2}) : \frac{π}{4}

arccos (\frac{1}{2})

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{4}

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}{4}

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

Combine the intervals

\frac{3 π}{4} + 2 πn < x < \frac{5 π}{4} + 2 πn or - \frac{π}{4} + 2 πn < x < \frac{π}{4} + 2 πn

Merge Overlapping Intervals

- \frac{π}{4} + 2 πn < x < \frac{π}{4} + 2 πn or \frac{3 π}{4} + 2 πn < x < \frac{5 π}{4} + 2 πn

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