How do you find the exact value of #cos2theta-cos^2theta+costheta+1/4=0# in the interval #0<=theta<2pi#?

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Answer 1 · 2018-04-01T09:17:43

#color(blue)(pi/3 , (5pi)/3#

Identity:

#color(red)bb(cos(2x)=2cos^2(x)-1)#

Substitute this into the equation:

#2cos^2(theta)-1-cos^2(theta)+cos(theta)+1/4=0#

Simplify:

#cos^2(theta)+cos(theta)-3/4=0#

#4cos^2(theta)+4cos(theta)-3=0#

Let # \ \ \ \ u=cos(theta)#

Then:

#4u^2+4u-3=0#

Factor:

#(2u+3)(2u-1)=0=>u=-3/2 and u=1/2#

But:

#u=cos(theta)#

#cos(theta)=1/2#

#cos(theta)=-3/2# *

#color(red)(-1<=cos(theta)<=1#

#cos(theta)=1/2#

#theta=arccos(cos(theta))=arccos(1/2)=>theta=color(blue)(pi/3 , (5pi)/3#