SOLUTION: Find all solutions in degress in the interval [0,360] 6cos^2x+7cosx+2=0 Thank you friends

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Question 1101265: Find all solutions in degress in the interval [0,360]
6cos^2x+7cosx+2=0
Thank you friends

Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
It's a quadratic in cos(x)
Check for extraneous solutions.

Answer by ikleyn(52898) About Me  (Show Source):
You can put this solution on YOUR website!
.
Instruction from Alan is correct, but I think that for the person who posted it, it is not enough.


So I put the entire solution here.


Introduce new variable u = cos(x). Then your equation takes the form

6u^2 + 7u + 2 = 0.


It is QUADRATIC EQUATION for the unknown "u". To solve it, apply the quadratic formula

u%5B1%2C2%5D = %28-7+%2B-+sqrt%287%5E2+-4%2A6%2A2%29%29%2F%282%2A6%29 = %28-7+%2B-+1%29%2F12.


You have two solutions for u:


1)  u%5B1%5D = %28-7%2B1%29%2F12 = -6%2F12 = -1%2F2.

    It leads us to equation cos(x) = -1%2F2 which has TWO solutions x = 120 degs and/or x = 240 degs in the given interval.


2)  u%5B2%5D = %28-7-1%29%2F12 = -8%2F12 = -2%2F3.

     It leads to equation cos(x) = -2%2F3 which has TWO solutions x = arccos(-2/3) degs and/or x = arccos(-2/3) degs + 180 degs  in the given interval.


Answer.  The given equation has 4 (four) solutions:

         x= 120 degs;  x= 240 degs;  x= arccos(-2/3) degs  and  x = arccos(-2/3) degs + 180 degs.

Solved.