Question

How do you solve for all solutions of x in `sin2x-1=0`?

Answer 1

(This is probably an attempt to try to trick you into using a double angle formula).

Let `theta = 2x`
So the problem becomes:
`sin(theta) - 1 = 0`
or
`sin(theta) = 1`

Within the range [`0, 2Pi`]
`theta = Pi/2` or `(3Pi)/2`
So (within the range [`0, 2Pi`]
`x = Pi/4` or `3Pi/4`

These solutions are repeated for every cycle of `2Pi`
so `x epsilon { Pi/4 + n2Pi, (3Pi)/4 + n2Pi }`
for all integer values of `n`