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

1 Answer
Oct 26, 2015

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