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Unformatted text preview: gardless. Second, as we have mentioned, there is no universally accepted range of the
arcsecant function. For that reason, we adopt the advice given in Section 10.3 and convert
this to the cosine problem cos(x) = − 5 . Adopting an angle approach, we consider the
equation cos(θ) = − 3 and note our solutions lie in Quadrants II and III. The reference
angle α satisﬁes 0 < α < π with cos(α) = 3 . We look to the arccosine function for help.
The real number t = arccos 5 satisﬁes 0 < t < π and cos(t) = 5 , so our reference angle
is α = arccos 3 radians. Proceeding as before, we ﬁnd the Quadrant II solutions to be
θ = π − α + 2πk = π − arccos 3 + 2πk for integers k . In Quadrant III, one angle with
reference angle α is π + α, so our solutions here are θ = π + α + 2πk = π + arccos 5 + 2πk for
12 When in doubt, write them out! 10.6 The Inverse Trigonometric Functions 719 integers k . Passing back to real numbers, we state our solutions as t = π − arccos
or t = π + arccos 3 + 2πk for integers k .
y + 2πk y 1 3
5 1 α
1 x α 1 x It is natural to wonder if it is possible to...
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