SOLUTION: I have one more question, my teacher told me that I need to factor the following problem in order to find the solutions but I'm not sure how to do it. Here's the question: Find

Question 936244: I have one more question, my teacher told me that I need to factor the following problem in order to find the solutions but I'm not sure how to do it. Here's the question:
Find all solutions to the equation in the interval [0,2pi). 2cos^2x + cosx-1=0.
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
Factoring wanted:
2cos^2x + cosx-1=0
and what you wrote will not render on the system:
AMP Parsing Error of [2cos^2x + cosx-1=0]: Invalid function '\x+cosx-1=0': opening bracket expected at /home/ichudov/project_locations/algebra.com/templates/Algebra/Expression.pm line 70. .

Assuming to mean, .

Substituting for , you can express as
.
Is this factorable?
You can try logical combinations or resort to finding discriminant.
'
Discriminant:, and this is NOT zero.
A discriminant of zero would indicate the quadratic expression factorable.

So far, you seem correct, in that your expression given does not seem factorable.

Continue with general solution of a quadratic equation.
Still in terms of u,

OR , meaning the roots are found, so you can give the equation as:
.

THAT means, when reversing the substitution,
.

Finally, your teacher is correct. The left member CAN be factored, but using knowledge of the general solution for a quadratic formula was very helpful in finding that factorization.
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