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**Unformatted text preview: **lutions lie in [0, 2π ), we substitute integer values for k . The solutions we keep come
from the values of k = 0 and k = 1 and are x = 5π , 7π , 17π and 19π . To conﬁrm these
12 12
12
12
√ answers graphically, we plot y = cos(2x) and y = − 23 over [0, 2π ) and examine where these
two graphs intersect. We see that the x-coordinates of the intersection points correspond to
our exact answers.
2. As we saw in Section 10.3, equations involving cosecant are usually best handled by converting
√
√
1
1
the cosecants to sines. Hence, we rewrite csc 1 x − π = 2 as sin 3 x − π = √2 = 22 .
3
√ π
There are two values in [0, 2π ) with sine 22 : π and 34 . Since the period of sine is 2π , we get
4
π
1
3π
1
3 x − π = 4 +2πk or 3 x − π = 4 +2πk for integers k . Solving for x, we get our general solution
x = 15π + 6πk or x = 21π + 6πk for integers k . Checking these answers, we get that for any
4
4
√
1
π
integer k , csc 3 15π + 6πk − π = csc 54 + 2πk − π = csc...

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