Question #a43bd

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Answer 1 · 2016-10-11T10:03:36

$x = pi$

Notice, before proceeding, that we must have $cosx != 0$ due to the presence of $2/cosx$ in the equation.

$cosx-1 = 2/cosx$

$=> cosx(cosx-1) = cosx(2/cosx)$

$=> cos^2x - cosx = 2$

$=> cos^2x - cosx - 2 = 0$

$=> (cosx+1)(cosx-2) = 0$

$=> cosx = -1$ or $cosx = 2$

As $|cosx|<=1$ for $x in RR$ , $cosx=2$ has no real solutions. Thus we are left with $cosx=-1$ as the only option.

$cosx = -1$

$=> x = pi+2pin, n in ZZ$ .

As we have the restriction $x in (0, 2pi)$ , the only choice for $n$ which works is $n=0$ . Thus we get our solution:

$x = pi$

Checking, we find that it satisfies the given equation and does not violate the restriction of $cosx != 0$ we set in the beginning, as desired.