```Question 610291
{{{tan(x)*sin^2(x) = 2tan(x)}}}
The only way to use factoring to solve an equation is when the factored expression is equal to zero. So before we start factoring we need a zero on one side. Subtracting 2tan(x) from each side we get:
{{{tan(x)*sin^2(x) - 2tan(x) = 0}}}<br>
Now we can factor. The common factor is tan(x):
{{{tan(x)*(sin^2(x) - 2) = 0}}}
This is all the factoring we can do.<br>
From the Zero Product Property we know that for a product to be zero, like this one, one (or more) of the factors must be zero. So:
tan(x) = 0 or {{{sin^2(x) - 2 = 0}}}
First let's solve the second equation:
{{{sin^2(x) - 2 = 0}}}
{{{sin^2(x) = 2}}}
Find the square root of each side:
{{{sin(x) = sqrt(2)}}} or {{{sin(x) = -sqrt(2)}}}
Since both {{{sqrt(2)}}} and {{{-sqrt(2)}}} are not possible values for sin (one is larger than 1 and the other is smaller than -1), there is no solution to {{{sin^2(x) = 2}}}.<br>
So the only solutions will come from:
tan(x) = 0
tan is zero at 0 and at {{{pi}}} and at all other angles that are coterminal with one of these two. The syntax for this is:
{{{x = 0 + 2*pi*n}}}
or
{{{x = pi + 2*pi*n}}}
This is the general solution to your equation.<br>
Last we use the general solution to find all the specific solutions in the specified interval: {{{0 <= x <= 2pi}}}. The way to do this is to replace the "n" in the general solution equations with various integers until you have found all the x's that are in the interval. If we try n = 0 in the first equation we get x = 0, which is in the interval. If we try n = 0 in the second equation we get {{{x = pi}}}, which is in the interval. If we try n = 1 in the first equation we get {{{x = 2pi}}}, which is in the interval. As you try other n's (and don't forget to try negative integers) in either equation you should find that the x's you get are either below 0 or above {{{2pi}}}. So the only solutions in the specified interval are:
0, {{{pi}}} and {{{2pi}}}<br>
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P.S. In response to your questions...
0 and {{{2pi}}} are co-terminal angles (i.e their terminal sides are the same). So they will have all the same values for sin, cos, tan, etc. But 0 and {{{2pi}}} are not themselves the same number/angle. So they both count separately as solutions to your equation. {{{-2pi}}}, {{{4pi}}}, {{{-4pi}}} and many, many others are co-terminal with 0, too. And they fit your equation, too. But they are not in the specified interval.<br>
Because the Trig functions are periodic, there are literally an infinite number of solutions to equations like these. Obviously we cannot list an infinite number of solutions. the "+{{{2pi*n}}}" (some books use "k" instead of "n") is a way for us to say "or any other co-terminal angle".
So
{{{x = 0 + 2*pi*n}}}
says
"x is 0 or any angle that is co-terminal with 0."
and
{{{x = pi + 2*pi*n}}}
says
"x is {{{pi}}} or any angle that is co-terminal with {{{pi}}}."
By replacing the n's (or k's) in these equations with various integers we get specific values for x that are solutions to your equation. Since there are an infinite number of integers, the "+{{{2pi*n}}}" is a clever way for us to the infinity of solutions without having to list an infinity of numbers.```