How do you solve #2cos^2 x + sin x - 2 = 0# over the interval 0 to 2pi?

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1 Answer
sente
Feb 10, 2016

The set of solutions to #2cos^2(x)+sin(x)-2 = 0# is #{0, pi/6, (5pi)/6, pi, 2pi}#

Explanation:

Using the identity #sin^2(x) + cos^2(x) = 1# we have

#2cos^2(x)+sin(x)-2 = 2(1-sin^2(x))+sin(x)-2#

#=2-2sin^2(x)+sin(x)-2#

#=-2sin^2(x)+sin(x)#

#=sin(x)(-2sin(x)+1) = 0#

Thus the equation is true when either
#sin(x) = 0#
or
#-2sin(x)+1 = 0 <=> sin(x) = 1/2#

On the interval #[0, 2pi]# we have #sin(x) = 0# if and only if #x in {0, pi, 2pi}#

On the interval #[0, 2pi]# we have #sin(x) = 1/2# if and only if #x in {pi/6, (5pi)/6}#

Thus the set of solutions to #2cos^2(x)+sin(x)-2 = 0# is #{0, pi/6, (5pi)/6, pi, 2pi}#

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