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

1 Answer
Trevor Ryan.
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#

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