2cos^2x-3cosx+1=0, finding x?

Question

I have NO idea what to do. Help? Step by step would be greatly appreciated, but just giving me an idea where to start off would be wonderful too. All I can figure out is that there's supposed to be more than one solution for x... -helpless shrug-

Oh, and it *is* -3cosx+1, not -3cos(x+1). I don't know if that makes any difference, but I thought I'd go ahead and mention that. (Yes, I'm awful at math.)

Answer 1

Yes that makes a huge difference



let cosx= y



then the equation becomes



2y^2-3y+1=0



2y^2-2y-y+1=0



2y(y-1)-1(y-1)=0



(2y-1)(y-1)=0



y= 1/2 or y=1



now substitute back



cosx= 1/2 or cosx= 1



therefore cos 60= 1/2

and cos0= 1

Answer 2

Let y = cos(x)



Your equation now becomes

2y^2 - 3y + 1 = 0



and now you will recognize that this is a simple quadratic equation and can be factored as:

(2y-1)(y-1) = 0



So x has solutions y =1 or y = 1/2



Now y = cos(x) so that means

cos(x) = 1 or cos(x) = 1/2



Let's look at cos(x) = 1. We know that cos(0) = 1 and it is periodic with period 2π So for y = 1 we have x = 0 + 2nπ where n is any integer.



Now let's look at cos(x) = 1/2. We know that cos(π/3) = 1/2 so we now have:

x = π/3 + 2nπ



However, we also know that cos(-π/3) = 1/2 [remember that the cosine is an even function]



so we also have a second set of solutions: x = -π/3 + 2nπ

Answer 3

You need to factor the quadratic. Let's make a substitution, because I think that that will make it clearer what's going on. Let z = cos(x). We can rewrite the equation as

2z^2 - 3z + 1 = 0

Now you can factor this like any old quadratic.

(2z - 1)(z - 1) = 0

z = 1, 1/2

Now, we have an equation for each value of z:

1 = cos(x) and 1/2 = cos(x)

In the range [0, 2pi], we have

x = 0, pi/3, 5pi/3, 2pi

Answer 4

2 * (cos(2x)) + 3 * cos(x) + a million = 0 2 * (cos(x)^2 - sin(x)^2) + 3cos(x) + a million = 0 2cos(x)^2 - 2sin(x)^2 + 3cos(x) + a million = 0 2cos(x)^2 - 2 * (a million - cos(x)^2) + 3cos(x) + a million = 0 2cos(x)^2 - 2 + 2cos(x)^2 + 3cos(x) + a million = 0 4cos(x)^2 + 3cos(x) - a million = 0 cos(x) = u 4u^2 + 3u - a million = 0 u = (-3 +/- sqrt(9 + sixteen)) / (2 * 4) u = (-3 +/- 5) / 8 u = -8/8 , 2/8 u = -a million , a million/4 cos(x) = -a million x = 3pi/2 cos(x) = a million/4 x = +/- seventy 5.5225 * pi / a hundred and eighty

Answer 5

Treat cos(x) as a variable all on its own...

Call it y for the moment, and your equation becomes:

2y^2 - 3y + 1 = 0

use quadratics to solve for y:

(2y - 1)(y - 1) = 0

y = 1/2 or y = 1

so cos(x) = 1/2 or cos(x) = 1

if cos(x) = 1 then x = 0 + 2kpi

if cos(x) = 1/2 then x = +/- pi/3 + 2kpi

Answer 6

Firstly, say n=cos(x):



2n^2-3n+1=0



(2n-1)(n-1)=0



n=1, 1/2



Put cos(x) back in for n:



cos(x)=1, 1/2



x=0, 60 (deg)

Answer 7

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