posted by Rae on .
Please help I need to solve the following question for x, where 0<x<2pi. and then write a general solution for the equation. The question is 2cos^2 x + cos x =1. I need to solve this using radian measurement. I solved it using algebra and was told I wasn't allowed. Could you please show me the steps to solve it using radian measurement and then how to write a general equation also using radian measurement. Please and thanks
there is two ways you can do this...
find the common factor of 2cos^2x+cosx=1
and bring out the front which = cosx(2cosx+1)=1
therefore either cosx=1 OR 2cosx+1=1 (because there will be multiple answers)
for cosx=1 use the unit circle to show where cosx=1 which is on the x axis because cos(pi)=1 which is show at points 0 and pi on the unit circle.
for 2cosx+1=1 simplify to cosx=-1/2
cos(pi/3)=-1/2 therefore x=pi/3
graph this in the 2nd and 3rd quadrant (s and t) because cos is negative in those quadrants
you should now have 4 "lines" on your unit circle graph. start at zero going anti clock wise labelling the points.
which will be x=0, (pi- pi/3), pi, (pi+pi/3) which simplified will equal
0, 2pi/3, pi, 4pi/3 these are your four possible answers for x.
2/ the other way to do this is using the quadratic formula.
in the equation 2cos^2x+cosx=1 , let cosx=t and substitute that in
so your equation should now be 2t^2+t=1
take the 1 over to the other side so it is 2t^2+t-1=0 which gives you a quadratic put this into the quadratic formula
so it will be now t= 1 +or- the square root (1^2-4*2*-1)all divided by 2*2
which simplifies to t=(1+or- 3)/4
therefore t= 1 OR -1/2
sub cosx=t into that
therefore cosx=1 OR cosx=-1/2
its the the same as 1/ from now on..
hope this helps.:D
I disagree with "belle"
2cos^2 x + cos x =1
2cos^2 x + cos x -1 = 0
(2cosx - 1)(cosx + 1) = 0
cosx = 1/2 or cosx = -1
case1: cosx = 1/2
we know cosπ/6 = 1/2 and the cosine is positive in I and IV
so x = π/6 or x = 2π - π/6 = 11π/6
case 2: cosx = -1
x = π
so x = π/6 , π, 11π/6
since the period of cosx is 2π, general solutions would be
π/6 + 2kπ
π + 2kπ or π(1+2k)
11π/6 + 2kπ, where k is an integer.