Math help?

Question

If i had sin2x(sin3x+cos3x)=2 how would i solve for x.
similarly if i had sin18x(sin6X+cos2x)
or sin14x(Sin6x+cos4x)

Answer 1

The trick in this case is that sin and cos are always between -1 and 1.
So if sin 2x was 0.9, say, the biggest sin2x(sin3x+cos3x) could be would be 0.9(1+1) = 1.8.
The only way to get 2 is to make sin 2x = sin 3x = cos 3x = 1 or -1.

However, its impossible for sin 3x = cos 3x = 1 or -1 (they never are both 1 or -1 at the same time).

So theres actually no solutions.

You can apply similar reasoning to the others - since you know each are 1 or -1, there are very limited possibilities for x.

Answer 2

I think maybe you need to divide the x and then distribute. Plug it in to see if you get the right answer. I am just thinking off the top of my head. Hope that helps.