How do you solve #sin3x=cos3x*tan3x# for x in the interval [0,2pi)?

1 Answer
Alan P.
Apr 26, 2016

The given equation is true for all values in the range #[0,2pi)#

Explanation:

You could (temporarily) replace #3x# with #theta#
so the given equation would look like:
#color(white)("XXX")sin(theta)=cos(theta)*tan(theta)#

But we know that #tan(theta)=(sin(theta))/(cos(theta))#

So we have that the given equation is equivalent to
#color(white)("XXX")sin(theta)=(cancel(cos(theta)))/color(white)("X")*(sin(theta))/(cancel(cos(theta)))#

Which is true for all values of #theta# and therefore true for all values of #x#.

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