As a general description, there are 3 steps. These steps may be very challenging, or even impossible, depending on the equation.

Step 1: Find the trigonometric values need to be to solve the equation.

Step 2: Find all 'angles' that give us these values from step 1.

Step 3: Find the values of the unknown that will result in angles that we got in step 2.

**(Long) Example**

Solve: #2sin(4x- pi/3)=1#

Step 1: The only trig function in this equation is #sin#.

Sometimes it is helpful to make things look simpler by replacing, like this:

Replace #sin(4x- pi/3)# by the single letter #S#. Now we need to find #S# to make #2S=1#. Simple! Make #S=1/2#

So a solution will need to make #sin(4x- pi/3)=1/2#

Step 2: The 'angle' in this equation is #(4x- pi/3)#. For the moment, let's call that #theta#. We need #sin theta = 1/2#

There are infinitely many such #theta#, we need to find them all.

Every #theta# that makes #sin theta = 1/2# is coterminal with either #pi/6# or with #(5 pi)/6#. (Go through one period of the graph, or once around the unit circle.)

So #theta# Which, remember is our short way of writing #4x- pi/3# must be of the form: #theta = pi/6+2 pi k# for some integer #k# or of the form #theta = (5 pi)/6 +2 pi k# for some integer #k#.

Step 3:

Replacing #theta# in the last bit of step 2, we see that we need one of: #4x- pi/3 = pi/6+2 pi k# for integer #k#

or #4x- pi/3 = (5 pi)/6+2 pi k# for integer #k#.

Adding # pi/3# in the form #(2 pi)/6# to both sides of these equations gives us:

#4x = (3 pi)/6+2 pi k = pi/2+2 pi k# for integer #k# or

#4x = (7 pi)/6+2 pi k# for integer #k#.

Dividing by #4# (multiplying by #1/4#) gets us to:

#x= pi/8+(2pi k)/4# or

#x=(7 pi)/24+(2 pi k)/4# for integer #k#.

We can write this in simpler form:

#x= pi/8+pi/2 k# or

#x=(7 pi)/24+pi/2 k# for integer #k#.

**Final note** The Integer #k# could be a positive or negative whole number or 0. If #k# is negative, we're actually subtracting from the basic solution.