Asked by Rae
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
Answers
Answered by
Belle
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
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
Answered by
Reiny
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.