SOLUTION: 3cos^2 thetha+2cos theta = 1 interval [0deg,360deg)

Algebra ->  Trigonometry-basics -> SOLUTION: 3cos^2 thetha+2cos theta = 1 interval [0deg,360deg)       Log On


   

Question 1203221: 3cos^2 thetha+2cos theta = 1 interval [0deg,360deg)

Found 3 solutions by MathLover1, ikleyn, math_tutor2020:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
+3cos%5E2+%28thetha%29%2B2cos%28+theta%29+=+1+interval [+0+%3Ctheta+%3C+360)

let +cos%28+theta%29+=u

+3u%5E2%2B2u=1
+3u%5E2%2B2u-1=0
+3u%5E2%2B3u-u-1=0
+%283u%5E2-u%29%2B%283u-1%29=0
+%28u+%2B+1%29+%283+u+-+1%29+=+0
+u=-1 =>+cos%28+theta%29+=-1
or
+u=1%2F3 =>+cos%28+theta%29+=1%2F3


general solutions:

+theta+=cos%5E-1%28+-1%29:
+theta+=+2pi%2A+n+-+pi,++n element++Z
+theta++=+2pi%2A+n+%2B+pi, +n+element +Z

+theta+=cos%5E-1%28+1%2F3%29:

+theta+=+2pi%2A+n+-+cos%5E%28-1%29%281%2F3%29, +n element +Z
+theta++=+2pi%2A+n+%2B+cos%5E%28-1%29%281%2F3%29, +n+element++Z

Real solutions in interval [+0+%3Ctheta+%3C+360):
+theta++=+pi
+theta++=+3pi
+theta++=+5pi
+theta+=+7pi
+theta++=+9pi

solutions
+theta++=+2pi%2An-cos%5E%28-1%29%281%2F3%29, 1%3C=n%3C=57
+theta++=+2pi%2An%2Bcos%5E%28-1%29%281%2F3%29, 0%3C=n%3C=57

Answer by ikleyn(52884) About Me  (Show Source):
You can put this solution on YOUR website!
.
3cos^2 thetha+2cos theta = 1 interval [0deg,360deg)
~~~~~~~~~~~~~~~~~~~~


        The answer presented in the post by @MathLover1 is absurdist and incorrect.
        I came to bring you a correct answer together with a correct solution. 


Let cos(theta) = u. Then the original equation is

    3u^2 + 2u = 1.


Transform it equivalently

    3u^2 + 2u - 1 = 0,

    (3u-1)*(u+1) = 0     <<<---=== by factoring.


So, the quadratic equation has two solutions: u= -1 and u= 1/3.


If u= -1, it means cos%28theta%29 = -1, hence, theta = 180 degrees, in the assigned interval.


If  u= 1/3,  it means  cos%28theta%29 = 1/3.  Then there are two solutions for theta  in the assigned interval  

    theta = arccos(1/3) = 70.53 degrees  and  theta = 360-70.53 = 289.47 degrees.


ANSWER.  The solutions for  theta  in the assigned interval are 180 degs,  70.53 degs  and  289.47 degs.

Solved.



Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Answers:
theta = 70.529 (approximate)
theta = 180 (exact)
theta = 289.471 (approximate)
Round the decimal values however your teacher instructs.


Explanation

I'll use x in place of theta (symbol theta)
The reason for this is to allow us to graph on the xy grid.

The equation to solve is
3cos%5E2%28x%29%2B2cos%28x%29+=+1
which can be rearranged to
3cos%5E2%28x%29%2B2cos%28x%29+-+1+=+0

Let
f%28x%29+=+3cos%5E2%28x%29%2B2cos%28x%29+-+1
The goal is to find the roots or x-intercepts.

Open up your favorite graphing app.
I'll go with Desmos. GeoGebra is another good option.
A calculator like a TI83 or TI84 works also.

We will need Desmos to be in degrees mode. Click the wrench icon in the upper right corner. Then swap from "radians" to "degrees".

In Desmos, we'll type in f(x) = 3cos^2(x)+2cos(x) - 1
To restrict the domain, append {0 <= x < 360}

The full input is f(x) = 3cos^2(x)+2cos(x) - 1 {0 <= x < 360}

Check out this link to see what I mean
https://www.desmos.com/calculator/dbjhvzx0cg
Refer to the first box in the left-hand panel.

The curve looks like a strange W of sorts.
It has three x-intercepts
x = 70.529 (approximate)
x = 180 (exact)
x = 289.471 (approximate)
Those are the 3 solutions to the equation.


Side notes:
  • Unfortunately Desmos appears to only allow accuracy up to 3 decimal places. Use GeoGebra for better precision.
  • WolframAlpha is another good solver to use. There are plenty others.
  • The graph window in Desmos is: xmin = -50, xmax = 400, ymin = -6, ymax = 6
  • In Desmos you can click on the x intercept location to have its coordinates show up