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Popular >

sin(2x)>= 0\land cos(x)>0

Solution

sin (2 x) \geq 0 and cos (x) > 0

Solution

πn \leq x < \frac{π}{2} + πn

Interval Notation

[πn, \frac{π}{2} + πn)

Decimal Notation

πn \leq x < 1.57079 \dots + πn

Solution steps

sin (2 x) \geq 0 and cos (x) > 0

sin (2 x) \geq 0 : πn \leq x \leq \frac{π}{2} + πn

sin (2 x) \geq 0

For

sin (x) \geq a

, if

- 1 < a < 1

then

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

arcsin (0) + 2 πn \leq 2 x \leq π - arcsin (0) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

arcsin (0) + 2 πn \leq 2 x and 2 x \leq π - arcsin (0) + 2 πn

arcsin (0) + 2 πn \leq 2 x : x \geq πn

arcsin (0) + 2 πn \leq 2 x

Switch sides

2 x \geq arcsin (0) + 2 πn

Simplify

arcsin (0) + 2 πn : 2 πn

arcsin (0) + 2 πn

Use the following trivial identity:

arcsin (0) = 0

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}

= 0 + 2 πn

0 + 2 πn = 2 πn

= 2 πn

2 x \geq 2 πn

Divide both sides by

2

2 x \geq 2 πn

Divide both sides by

2

\frac{2 x}{2} \geq \frac{2 πn}{2}

Simplify

x \geq πn

x \geq πn

2 x \leq π - arcsin (0) + 2 πn : x \leq \frac{π}{2} + πn

2 x \leq π - arcsin (0) + 2 πn

Simplify

π - arcsin (0) + 2 πn : π + 2 πn

π - arcsin (0) + 2 πn

Use the following trivial identity:

arcsin (0) = 0

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 - 0 + 2 πn

π - 0 + 2 πn = π + 2 πn

= π + 2 πn

2 x \leq π + 2 πn

Divide both sides by

2

2 x \leq π + 2 πn

Divide both sides by

2

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

Simplify

x \leq \frac{π}{2} + πn

x \leq \frac{π}{2} + πn

Combine the intervals

x \geq πn and x \leq \frac{π}{2} + πn

Merge Overlapping Intervals

πn \leq x \leq \frac{π}{2} + πn

cos (x) > 0 : - \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

cos (x) > 0

For

cos (x) > a

, if

- 1 \leq a < 1

then

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

- arccos (0) + 2 πn < x < arccos (0) + 2 πn

Simplify

- arccos (0) : - \frac{π}{2}

- arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

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}{2}

Simplify

arccos (0) : \frac{π}{2}

arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

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}{2}

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

Combine the intervals

πn \leq x \leq \frac{π}{2} + πn and - \frac{π}{2} + 2 πn < x < \frac{π}{2} + 2 πn

Merge Overlapping Intervals

πn \leq x < \frac{π}{2} + πn

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