How can I find all the solutions to cos(3x) = sqrt(2)/2 for 0<=x<=2pi ?

Finding the solutions to trig equations begins in a very similar way to when we do it at GCSE. We start with:cos(2x) = 2^-1/2 0 <= x <=piand correct our range, as we are interested in 2x not just x: 0 <= 2x <= 2pi .Then we find our first solution 2x = cos^-1(2^-1/2) = pi/4So we have one solution: 2x = pi/4. Our next solution comes from looking at the graph of cos(x), which is symmetrical around the y-axis. From this we can see that 2x = -pi/4 will also be a solution. Then we can add 2pi to each solution until we leave our range. This gives us a set of answers: 2x = pi/4, 7pi/4. We don't need -pi/4 or 9pi/4 because they are not in our range.So x = pi/8, 7pi/8.