This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Next, we recall that for the square root to be deﬁned, we
need 1 − cot(x) ≥ 0. Unlike the inequalities we solved in Example 10.7.3, we are not restricted
here to a given interval. For this reason, we employ the same technique as when we solved
equations involving cotangent. That is, we solve the inequality on (0, π ) and then add integer
multiples of the period, in this case, π . We let g (x) = 1 − cot(x) and set about making a sign
diagram for g over the interval (0, π ) to ﬁnd where g (x) > 0. We note that g is undeﬁned for
x = πk for integers k , in particular, at the endpoints of our interval x = 0 and x = π . Next,
we look for the zeros of g . Solving g (x) = 0, we get cot(x) = 1 or x = π + πk for integers k
and only one of these, x = π , lies in (0, π ). Choosing the test values x = π and x = π , we
get g π = 1 − 3, and g π = 1.
(−) 0 (+)
We ﬁnd g (x) > 0 on π
4,π π . Adding multiples of the period we get our solution to c...
View Full Document