Solve 2cos^2 x + cos x -1 = 0?

Question

What I have so far is:
(2cosx -1)(cosx +1)
2cosx -1=0
2cosx=1
cosx=1/2

cosx +1=0
cosx =-1
I just don't know where to go from there...anybody?

Answer 1

now you have finished half of the problem. you just need to find the values of x from here.
the general formula for a cosine equation:
if cosx=cosy, x=2*n*pi+y
so, the first set of values are cosx=1/2
cosx=cos(pi/3)
so, x=2*n*pi+pi/3

now, for the second root, cosx=-1
cosx=cos(pi)
so, x=2*n*pi+pi


this is the complete solution for ur question

Answer 2

(2cosx -1)(cosx +1)
2cosx -1=0
2cosx=1
cosx=1/2

cosx +1=0
cosx =-1

cosx = 1/2 => cosx=cos (pi/3) => x = +- pi/3 + k*2*pi (k is an integer)

cosx = -1 => x = pi + k *2pi (k is an integer)

Answer 3

You have done a good job so far.

Now draw your unit circle and we will work in radians. Below k represents any integer. From the circle we see,

cos x=1/2 if x=Pi/3+2 k Pi or x=5Pi/3+2k Pi

cos x=-1 if x =Pi+2 k Pi

Use should try to do the same problem usine sine instead of cosine. :)

Good luck!

Answer 4

when u got,,,
cosx = 1/2,,
the answer is--
60,,and 360-60=300
soo,,60 and 300 are one set of answers,,

when u got,,,
cosx = -1,,
the answer is--
180,,only,,(as,,360-180=180)

therefore,,the answers=
60,,300,,180,,

theres a formula to be used--
cosx=cos(360-x)

Answer 5

Looks good. Now use a graphing calculator and plot these equations. You should be able to find values for X. Make sure to use radians. Good luck.

Answer 6

cos[x] + cos{x} = 2 cos([x]+{x} / 2) cos([x]-{x} / 2) = 2 cos x cos ([x]-{x} / 2) 2 cos x cos ([x]-{x} / 2) = sin x =>cos ([x]-{x} / 2) = a million/2 tan x only one answer at x = forty 5

Answer 7

cosx=1/2
cosx=cospi/3
x=2npi+/-pi/3
cosx=-1
cosx=cospi
x=2npi+/-pi