When dealing with trigonometric functions such as sin, cos or tan, how do you solve the trigonometric equation when the argument of the function(s) is nx, where n is a real number not equal to 1.

In order to solve Trigonometric equations like these, you must consider how the argument of the function changes the period of the graph. There are many methods of solving equations like this, but personally, I would begin by multiplying each endpoint of the domain by n, where n is the coefficient of x. You can then think of this as the 'new domain' of the equation is terms of nx, rather than just x. By doing this, you have essentially horizontally enlarged the graph by a scale factor of 1/n. I would then rearrange the equation to get the trigonometric function on one side, and then a constant on the other side, and would then proceed by solving the equation by using a CAST diagram or by using the periodic properties of the graph. The key point here is that you now have a 'new domain' and you must ensure that you find every solution for nx within this domain, rather than the original domain, which could be larger or smaller than the original domain, depending on the value of n. Once you have found every solution to the equation within the 'new domain', you can then divide each solution by n, in order to get all of the solutions in the original domain. Let me run through an example.
Example:Find all of the solutions to the trigonometric equation 8sin(2x) - 3 = 1, where 0<=x<=360.Firstly, I would establish the 'new domain' as 0<=2x<=720 by multiplying the domain by n, where n = 2 in this case.I would then rearrange the equation to get the trigonometric function on one side, and a constant on the other side, which gives sin(2x) = 1/2.I would then look to find all of the solutions to this equation within the 'new domain' by looking at the periodic property of the sin(x) graph.Clearly, the most obvious solution is 2x = 30. By considering the graph of sin(2x), you know that 2x = 150 is also a solution, since the graph is symmetric between 0<=2x<=180. There are no other solutions between 180<=2x<=360, since the graph takes negative y-values for these x-values. However, if 2x = 30 is a solution, then 2x = 390 must also be a solution by the periodic property of the graph. Likewise, if 2x = 150 is a solution, then 2x = 510 is also a solution. These are the only 4 solutions within the domain.Finally, since you now have 2x = 30, 150, 390, 510, you can then divide each solution by 2 and arrive at x = 15, 75, 195, 255