How do you find the general solutions for #2sin^2 β – 3sin β +2 = 0#?

1 Answer
Alan P.
Oct 19, 2015

#2sin^2(beta)-3sin(beta)+2 = 0#
has no solutions

Explanation:

Consider the equation
#color(white)("XXX")2x^2-3x+2=0#
#color(white)("XXXXXX")#where #x# is (temporarilly) used in place of #sin(beta)#

The determinant of #2x^2-3x+2#
is #Delta =(-3)^2-4(2)(2) = -7#
and we know
#color(white)("XXX")# if #Delta < 0# the quadratic has no Real roots.

If there are no Real solutions for #x#
#rArr# there are no Real solutions for #sin(beta)#
#rArr# there are no solutions for #beta# (since #sin# is a Real valued function).

We could also see this by considering the graph of
#2sin^2(beta)-3sin(beta)+2#
graph{2(sin(x))^2-3sin(x)+2 [-10, 10, -4.08, 5.92]}
Note that this expression does not touch the X-axis (i.e. it is never equal to #0#)

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