Hi Ben.
We had fun with this problem, in fact we solved it three ways!
Method 1: First we used the trig identities for sin(2x) and cos(2x), first writing cos(3x) = cos(x + 2x), and arrived at a quadratic in sin(x) which we solved to find sin(x).
Method 2: Then we realized that the subject line of your email was calculus, so although you might want to solve this problem using a formula we learned in first year calculus (it's quicker than using just trig identities anyway). Here's how that works out:
Abraham de Moivre was a Frenchborn mathematician who pioneered the development of analytic geometry and the theory of probability. This is the theory named for him:
cos(nx) + i sin (nx) = (cos x + i sin x)^{n } .
This means that for n=3 we have:
cos(3x) + i sin (3x) = (cos x + i sin x)^{3}
= cos^{3 } x + 3i cos^{2 } x sin x  3cos x sin^{2 } x  i sin^{3 } x
whereupon we can ignore the imaginary portions and leave the real part:
cos (3x) = cos^{3 } x  3cos x sin^{2 } x
Using the same reasoning, we can show that sin (2x) = 2 sin x cos x, which is a wellknown trig identity anway. Now since sin (2x) = cos (3x), we have:
2 sin x cos x = cos^{3 } x  3cos x sin^{2 } x
2 sin x = cos^{2 } x  3sin^{2 } x
and since sin^{2 } x + cos^{2 } x = 1,
2 sin x = (1  sin^{2 } x)  3sin^{2 } x
leaving
4sin^{2 } x + 2sin x  1 = 0
which is just a quadratic equation that you can solve with the quadratic formula.
(This is the same quadratic we found using our first method.)
Method 3: Finally we realized that there is another trig identity you might use. It is that
cos(90^{o}  y) = sin(y)
Hence sin(2x) = cos(90^{o}  2x) and if sin(2x) = cos(3x) you have
cos(90^{o}  2x) = cos(3x)
So does this mean that 90^{o}  2x = 3x? In general this is not true because of the periodic nature of cosine, but in your situation the angle 3x is small and positive, 0 < 3x < 90^{o}, and hence in this particular case,
cos(90^{o}  2x) = cos(3x) does indeed imply that 90^{o}  2x = 3x.
Whatever method you use, be sure to check your answer(s) to make sure that 3x is a valid angle as described in the original question.
Hope this helps,
Stephen and Penny.
