Solving trigonometric equations

First rearrange to give sin(3x)=0.5

Now apply the sin^-1 function to both sides. The important thing to remember is that this is multi-valued - you can add 2nπ to the answer for any positive or negative integer n. (This is not such a strange concept - after all, when you take a square root you need to consider two answers, positive and negative).

For tan^-1 you add nπ, since it has a shorter cycle.

So this gives us that 3x = (π/6 or 5π/6) + 2nπ

Dividing by 3 gives the answer: x = (π/18 or 5π/18) + 2nπ/3

Depending on the range required for the answers, you may need to include several values of n.

Example 2: 4+sin x = 6 cos²x

Substitute 1-sin²x for cos²x

6sin²x + sin x -2=0

To make things clearer, it can help to change the variable. Let s=sin x;
Then 6s²+s-2=0
(2s-1)(3s+2)=0

And so, sin x=1/2 or -2/3.

To find x use the process of example 1.

It is also a good idea to count early in the process how many answers you expect to get. For sin, tan or cos = value, over one cycle, you get two answers. If it is sin 2x = value, over 2π, you get 4 values, etc.

There are specific ratios that you are expected to know. Remember two triangles: a) 1, 1, √2 which has 45° angles, and b) 1, √3, 2 which has 30° and 60° angles. You can think of a) as a square cut in half on the diagonal, and b) as an equilateral triangle cut in half.

If you remember these two triangles then you will easily be able to reconstruct sin, cos and tan of 30°, 45° and 60°.