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Popular >

cos(x)>0\land tan(x)>0

Solution

cos (x) > 0 and tan (x) > 0

Solution

2 πn < x < \frac{π}{2} + 2 πn

+2

Interval Notation

(2 πn, \frac{π}{2} + 2 πn)

Decimal Notation

2 πn < x < 1.57079 \dots + 2 πn

Solution steps

cos (x) > 0 and tan (x) > 0

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

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

tan (x) > 0

If

tan (x) > a

then

arctan (a) + πn < x < \frac{π}{2} + πn

arctan (0) + πn < x < \frac{π}{2} + πn

Simplify

arctan (0) : 0

arctan (0)

Use the following trivial identity:

arctan (0) = 0

x 0 \frac{3}{3} 13 arctan (x) 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} arctan (x) 0^{\circ} 3 0^{\circ} 4 5^{\circ} 6 0^{\circ}

= 0

0 + πn < x < \frac{π}{2} + πn

Simplify

πn < x < \frac{π}{2} + πn

Combine the intervals

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

Merge Overlapping Intervals

2 πn < x < \frac{π}{2} + 2 πn

Popular Examples

- 1 < cot (x) < 1

tan(θ)<0\land sec(θ)>0

tan (θ) < 0 and sec (θ) > 0

sin(θ)= 9/41 \land cos(θ)>0

sin (θ) = \frac{9}{41} and cos (θ) > 0

sin (x) 0 < x < 2 π

cos(-1)-pi<t<= pi

cos (- 1) - π < t \leq π