How do you solve trigonometric equations?

To help explain we will use the following example: Show that cos(2x) = 2cos²x - 1. Hence solve the equation cos(2x) + cosx = 0 for 0^o < x < 360^o. Answer: We know from our sum-difference formulas that cos(u+v) = cos(u)cos(v) - sin(u)sin(v) and in this case u = v = x. Simplifying to cos(2x) = cos²x - sin²x. We also know from our pythagorean identities that sin²x + cos²x = 1, implying that sin²x = 1 - cos²x. We can substitute this into our equation for cos(2x), giving cos(2x) = cos²x - (1 - cos²x). Expanding the brackets gives cos²x - 1 + cos²x. Simplifying gives 2cos²x - 1. This completes the first part of the question.  We are given the equation cos(2x) + cosx = 0 we can now substutute in cos(2x) = 2cos²x - 1 which gives 2cos²x - 1 + cosx = 0. Using the quadratic formula we find that cosx = (-1 + 3)/4 = 0.5 or cosx = (-1 - 3)/4 = -1. To find x we use inverse cos: x = cos^-1(0.5) = 60^o or x = cos^-1(-1) = 180^o. We can then sketch the cos(x) graph, 0^o < x < 360^o, find the points where x = 60^o and 180^o. We then draw vetricle lines across our sketch and find where these verticle lines touch the graph, this finds other solutions within the range, in this case there is one more solution, x = 300^o.