How do you solve $sec(x/2)=cos(x/2)$ over the interval 0 to 2pi?

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Answer 1 · 2016-02-23T22:05:38

$x=0$ and $x=2pi$

First, use the identity that $sec(x)=1/cos(x)$ to rewrite the equation:

$1/cos(x/2)=cos(x/2)$

Cross multiply, or multiply both sides by $cos(x/2)$ . This has the same function:

$1=cos^2(x/2)$

Take the square root of both sides.

$cos(x/2)=+-1$

Thus, we see that $x/2=0$ and $x/2=pi$ , since $cos(0)=1$ and $cos(pi)=-1$ .

$x/2=0" "=>" "x=0$

$x/2=pi" "=>" "x=2pi$

While there are an infinite number of solutions, these are the only two on the interval.

How do you solve sec(x/2)=cos(x/2)sec(x/2)=cos(x/2) over the interval 0 to 2pi?