Question #b623c

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Answer 1 · 2017-02-22T08:40:47

$theta = n pi$ where $n =$ all integers

Find a common denominator: $1/cos theta - cos^2 theta/cos theta = (1-cos^2 theta)/ cos theta$

Set the function $=0$

From the denominator: $cos theta = 0$ gives asymptotes

From the numerator you find the zeros:

$1-cos^2 theta = 0; cos^2 theta = 1; cos theta =+-1$

This occurs at $theta = ..., -2pi, -pi, 0, pi, 3pi, ...$

So $theta = n pi$ where $n =$ all integers