Question

How do you find all solutions of the equation `sin^2x=2sin x+3`?

Answer 1

First of all, let's analyze this equation. There is no restriction for `x` since the domain of a function `sin(x)` is all real numbers.

Secondly, we can introduce an intermediary variable `y=sin(x)` and express this equation in terms of `y`:
`y^2=2y+3` or
`y^2-2y-3=0`
This equation has solutions:
`y_(1,2)=(2+-sqrt(4+12))/2` or
`y_1=3` and `y_2=-1`

Now we have to take into consideration that `y=sin(x)` and, as such, is restricted in values from `-1` to `1` (inclusive).
Therefore, solution `y_1=3` would not lead to any value of `x`.
The second solution `y_2=-1` can be used to find `x`:
`sin(x)=-1` implies that
`x=-pi/2+2piN`, where `N` - any integer number or, in degrees,
`x=-90^o + 360^o*N`
This is a final solution to the problem.