Posted by **Rae** on Thursday, April 29, 2010 at 10:01pm.

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

- trig 30 -
**Belle**, Thursday, April 29, 2010 at 11:35pm
hey Rae,

there is two ways you can do this...

1/

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

- trig 30 -
**Reiny**, Thursday, April 29, 2010 at 11:59pm
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.

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