Find all the solutions of 2 cos 2x = 1 – 2 sinx in the interval 0 ≤ x ≤ 360°.

We know that having cos and sin functions in a question is going to cause us problems - we need a single function of x. So, step 1, we need to rewrite sin or cos in terms of the other. How do we choose which? Well, we can rewrite sin x as sqrt(1-cos²x), however this will leave us with a very messy function. So how about cos2x? We know that cos2x can be re-written in 3 ways, 2 of which, however, will leave us with the same problem of a combination sin and cos functions. Therefore, we choose to rewrite cos2x as 1 - 2sin²x. We now have 2 - 4sin²x = 1 - 2sinx. Rearranging this gives 4sin²x - 2sinx -1 = 0. We now have a quadratic inequality in sinx (we could let sinx = a here and solve 4a² -2a - 1 = 0) and can either solve this on our calulators or factorise it to give sinx = (1+sqrt(5))/4 and sinx = (1-sqrt(5))/4. These values are approximately equal to 0.8 and -0.3 and using a graph of sinx, we can find the x values corresponding with these sinx values. The graph shows there will be 2 x values for each sinx value in the given range, and these are 54^o and (180 - 54 =) 126^o for sinx = (1+sqrt(5))/4 and (-18 + 360 =) 342^o and (180 + 18 =) 198^o for sinx = (1-sqrt(5))/4.