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

1 Answer
May 14, 2015

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.