Unformatted text preview: − α, so the Quadrant IV solutions are θ = 2π − α + 2πk = 2π − arctan(2) + 2πk
for integers k . As we saw in Section 10.3, these solutions can be combined.12 One way to
describe all the solutions is θ = − arctan(2) + πk for integers k . Remembering that we are
solving for real numbers t and not angles θ measured in radians, we write our ﬁnal answer as
t = − arctan(2) + πk for integers k .
y y 1 1 α
1 x α 1 x Alternatively, we can forgo the ‘angle’ approach altogether and we note that tan(t) = −2
only once on its fundamental period − π , π . By deﬁnition, this happens at the value t =
arctan(−2). Since the period of tangent is π , we can capture all solutions by adding integer
multiples of π and get our solution t = arctan(−2) + πk for integers k .
3. The last equation we are asked to solve, sec(x) = − 5 , poses two immediate problems. First,
we are not told whether or not x represents an angle or a real number. We assume the
latter, but note that, once again, we will use angles and the Unit Circle to solve the equation
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