Trig help-urgent

Question

Solve the following inequality. Express the exact answer in interval notation, restricting your attention to 0 ≤ x ≤ 2𝜋.

tan(x) ≥ 1

Answer 1

Tangent is "opposite over adjacent". So to get a positive tangent, the opposite and adjacent would have to be the same sign: both positive or both negative.

Opposite and adjacent are the same sign (both positive) when the angle is in the First Quadrant (pointing to the upper right). Opposite and adjacent are the same sign (both negative) when the angle is in the Third Quadrant (pointing to the lower left). There are no other quadrants where the desired angle might lie.

Let's see where we've got a tangent greater than or equal to 1 in the First Quadrant. tan(x) = 1 when x = 45 degrees (i.e., when x = π/4 radians, i.e., exactly to the upper left) and the tangent gets bigger as x gets bigger. Of course, when x reaches 90 degrees (i.e., when x = π/2 radians), the angle is pointing straight up and the tangent becomes infinite (because the opposite side is infinite and the adjacent side is zero), so we don't want to let the angle grow all the way to 90 degrees. Therefore when x is in the First Quadrant, tan(x) is greater than or equal to 1 when x is in the interval [π/4, π/2).

Let's see where we've got a tangent greater than or equal to 1 in the Third Quadrant. tan(x) = 1 when x = 225 degrees (i.e., when x = 5π/4 radians, i.e., exactly to the lower left) and the tangent gets bigger as x gets bigger. Of course, when x reaches 270 degrees (i.e., when x = 3π/2 radians), the angle is pointing straight down and the tangent becomes negative infinity (because the opposite side is negative infinity and the adjacent side is zero), so we don't want to let the angle grow all the way to 270 degrees. Therefore when x is in the Third Quadrant, tan(x) is greater than or equal to 1 when x is in the interval [5π/4, 3π/2).

The problem asks us to find everywhere where tan(x)≥1, so we want both of these intervals.

So the answer is the union of these two intervals: [π/4, π/2) ∪ [5π/4, 3π/2). Thanks.

Answer 2

tanx >/=1

0</=x</=2pi

if tanx =1, x would = 45 degrees =pi/4, or 5pi/4 = 225 degrees

so those are lower boundaries of the intervals you want

[pi/4, pi/2), [5pi/4, 3pi/2)

or

[45, 90), [225,270) in degrees

two intervals in the 1st and 3rd quadrants

maybe you could call them the 2nd & 6th octants, if you divide the unit circle into 1/8's