How do you find all solutions of the equation in the interval #[0,2pi)# given #2cos^2x-cosx=1#?

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Answer 1 · 2017-02-23T09:54:24

The solutions are #S={0,2/3pi,4/3pi}#

We compare this equation to the quadratic equation

#ax^2+bx+c=0#

#2cos^2x-cosx-1=0#

We calculate the discriminant

#Delta=b^2-4ac#

#=(-1)^2-4*(2)*(-1)#

#=1+8=9#

As #Delta>0#, there are 2 real roots

#x=(-b+-sqrtDelta)/(2a)#

#cosx=(1+-sqrt9)/(4)#

#cosx=(1+3)/4=1# and #cosx=(1-3)/4=-1/2#

#x=0# and #x=2/3pi# and #x=4/3pi#