How do you solve #2sin^2x = 2 + cosx# in the interval 0 to 2pi?

1 Answer
MrsRudolph221
Mar 28, 2016

First, perform a Pythagorean substitution to remove the sine term from the left side: #2(1-cos^2(x))=2 + cos(x)# .

Explanation:

Simplify the left side : #2-2cos^2(x)=2+cos(x)#
Gather like terms and set equal to 0: #0=2cos^2(x)+cos(x)#
Factor the right side: #0=cos(x)(2cos(x) + 1)#
Use the Zero Product Property:
#cos(x) = 0# or #2cos(x)+1=0#
#cos(x)=0# or #2cos(x) = -1#
#cos(x)=0# or # cos(x) = -1/2#
So, #x = cos^-1(0)# or #x = cos^-1(-1/2)#
x = #pi/2# , #(3*pi)/2#, or #(2*pi)/3#, #(4*pi)/3#

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