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**Unformatted text preview: **r times.5 2. Analysis of sec2 (x) = tan(x) + 3 reveals two diﬀerent trigonometric functions, so an identity
is in order. Since sec2 (x) = 1 + tan2 (x), we get
sec2 (x)
1 + tan2 (x)
2 (x) − tan(x) − 2
tan
u2 − u − 2
(u + 1)(u − 2) =
=
=
=
= tan(x) + 3
tan(x) + 3 (Since sec2 (x) = 1 + tan2 (x).)
0
0
Let u = tan(x).
0 5
Note that we are not counting the point (2π, 0) in our solution set since x = 2π is not in the interval [0, 2π ). In
the forthcoming solutions, remember that while x = 2π may be a solution to the equation, it isn’t counted among
the solutions in [0, 2π ). 734 Foundations of Trigonometry
This gives u = −1 or u = 2. Since u = tan(x), we have tan(x) = −1 or tan(x) = 2. From
tan(x) = −1, we get x = − π + πk for integers k . To solve tan(x) = 2, we employ the
4
arctangent function and get x = arctan(2) + πk for integers k . From the ﬁrst set of solutions,
π
π
we get x = 34 and x = 54 as our answers which lie in [0, 2π ). Using the same sort of argument
we saw in Example 10.7.1, we get x = arctan(2) and x = π + arctan(2) as answers from our
second set of solutions which lie in [0, 2π ). Using a reciprocal identity, we rewrite the secant
1
as a cosine and graph y = (cos(x))2 and y = tan(x) + 3 to ﬁnd the x-values of the points...

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