Question 165742
<pre><font size = 4 color = "indigo"><b>

{{{cos(x)tan(x)=0}}}

We have to be careful here!  We might be tempted
to use the zero-factor principle. However that 
will lead us astray because {{{tan(x)}}} is not 
defined at ANY of the values of x for which
{{{cos(x)=0}}}!

So do NOT do this:

{{{cross(matrix(3,4, cos(x)=0,"","",tan(x)=0, 
               x=pi/2+n*pi,"","",x=0+n*pi,
                        "","","",x=n*pi))
 }}}

Instead use the identity {{{matrix(1, 3,tan(alpha), "=", sin(alpha)/cos(alpha))}}} to replace {{{tan(x)}}}.

{{{ matrix(1,3, cos(x)tan(x), "=", 0) }}}
{{{ matrix(1,5, cos(x), "*", sin(x)/cos(x), "=", 0)}}}

We can cancel {{{cos(x)}}}'s as long as we are careful 
not to allow any values of x which cause the 
denominator {{{cos(x)}}} to be zero.

{{{ matrix(1,5, cross(cos(x)), "*", sin(x)/cross(cos(x)), "=", 0)}}} 

That leaves:

{{{matrix(1,3,sin(x),"=",0)}}}

which has solution 

{{{x=n*pi}}}, where n is any integer.

And this is a valid solution since {{{cos(n*pi)}}}
is always either {{{""+1}}} or {{{-1}}}, and never {{{0}}}.

Edwin</pre>