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 0o < x < 360o. 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) = 60o or x = cos-1(-1) = 180o. We can then sketch the cos(x) graph, 0o < x < 360o, find the points where x = 60o and 180o. 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 = 300o.
