Question 1126640
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The product is 0 when either factor is 0.<br>
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skip these lines; I solved the wrong equation...<br>
The second factor is 0 when cos(2x) = -1; the solutions are
2x = 2pi/3 +/- 2k(pi)  -->  x = pi/3 +/- k(pi)
 or
2x = 4pi/3 +/- 2k(pi)  -->  x = 2pi/3 +/- k(pi)<br>
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here is the corrected discussion of the second factor...<br>
The second factor is 0 when 2cos(x) = -1, or cos(x) = -1/2.  The solutions are
2pi/3 +/- 2k(pi)
or
4pi/3 +/- 2k(pi)<br>
The first factor does not have exact zeros, the way the second one does.  Its zeros are when
tan(2x) = 16
2x = arctan(16) +/- k(pi)
x = (1/2)arctan(16) +/-(k/2)(pi)<br>
Use a calculator to get those zeros to the prescribed accuracy.