5.3 Continued- How to deal with more than just a typical x within a trig function?

An example of what this would look like is this: Find all solutions for x within the domain of [0,2π)
(cos2xcot2x)/(1-sin2x)=3

or

cos(x/2)-sin(x/2)=1

To solve problems like these we don't need to know complicated trig formulas such as half angle formulas. These aren't necessary in these problems because all the trig functions have the same angle, 2x or x/2.

The first step to solving problems like these is to have y equal the angle that you are trying to find. For instance y=2x or y=x/2. From here you should rewrite the problem so that it is now written with the y instead of the other angle.

The equations would now look like this:
(cosycoty)/(1-siny)=3

or

cosy-siny=1

From this point one should just simply solve the trig equations like any normal one for y.

Once this is done y will equal 5π/6 and π/6, 0 and π and 3π/2.

Now you should substitute the original angle back in for y. The equations would now look like this:

2x=5π/6 and 2x=π/6

or

x/2=0 and x/2=π and x/2=3π/2

From this point it is simple algebra to reach the answers that are 5π/12 and π/12, 0 and 2π and 3π

However as you may have noticed two of the answers from the second example, 2π and 3π, are not in the domain specified by the instructions. Therefore the only answer for that problem is 0.

Do you remember what the changing b does in the formula f(x)=a+sin(bx-c)+d?
Changing the value of b will change the period of the function.

Changing the value of b in the first problem causes there to be two more solutions. To find these you should add the 2π to the values of y. So now y=13π/6 and 17π/6. From here you can just solve the problem again as you did before, finding x to no be 13π/12 and 17π/12 as well.

This graph shows how the period can affect the amount of solutions there are within the domain permitted with in the problem.