Question #a43bd

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Answer 1 · 2016-10-11T10:03:36

$x = π$ $x = pi$

Notice, before proceeding, that we must have $cos x \neq 0$ $cosx != 0$ due to the presence of $\frac{2}{cos x}$ $2/cosx$ in the equation.

$cos x - 1 = \frac{2}{cos x}$ $cosx-1 = 2/cosx$

$\Rightarrow cos x (cos x - 1) = cos x (\frac{2}{cos x})$ $=> cosx(cosx-1) = cosx(2/cosx)$

$\Rightarrow {cos}^{2} x - cos x = 2$ $=> cos^2x - cosx = 2$

$\Rightarrow {cos}^{2} x - cos x - 2 = 0$ $=> cos^2x - cosx - 2 = 0$

$\Rightarrow (cos x + 1) (cos x - 2) = 0$ $=> (cosx+1)(cosx-2) = 0$

$\Rightarrow cos x = - 1$ $=> cosx = -1$ or $cos x = 2$ $cosx = 2$

As $| cos x | \leq 1$ $|cosx|<=1$ 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.