Question

2sin^2x-9sinx+7=0 What's the solution set?

Answer 1

`x=pi/2+2pik ; k in ZZ`

Explanation:

We may identify this as a quadratic equation in `sinx` and hence use the quadratic formula to find that

`sinx=(9+-sqrt(81-4xx2xx7))/(2xx2)=7/2 or 1`

Clearly `sinx=7/2` is impossible since `sinx in [-1;1] AA x in RR`
and so hence `sinx=1` is the only possibility.

`therefore x=sin^(-1)1=90^@=pi/2 rad`

But since the sine graph is repetitive, every full cycle (`2pi rad`)thereafter it will have the same value.
Hence the general solution is
`x=pi/2+2pik ; k in ZZ`