SOLUTION: Use a calculator to find all solutions in the interval (0, 2π). Round the answers to two decimal places. (Enter your answers as a comma-separated list. If there is no solution

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Question 691290: Use a calculator to find all solutions in the interval (0, 2π). Round the answers to two decimal places. (Enter your answers as a comma-separated list. If there is no solution, enter NO SOLUTION.)
5 cos3 x − 20 cos2 x + cos x − 4 = 0
Hint: Factor by grouping.

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
(In the future please use "^" to indicate exponents. Otherwise it is impossible to tell the difference between cos%5E3%28x%29 and cos(3x). I was not going to even try this problem because I could not tell the difference. It is only the "factor by grouping" that helped me figure out that the 3 and the 2 must be exponents.)

5cos%5E3%28x%29+-+20cos%5E2%28x%29+%2B+cos%28x%29+-+4+=+0
Factoring by grouping...
Group the first two terms and the last two terms:
%285cos%5E3%28x%29+-+20cos%5E2%28x%29%29+%2B+%28cos%28x%29+-+4%29+=+0
Factor out the Greatest Common Factor (GCF) of each group. (Note: This is one of the rare cases when one actually factors out a GCF of 1!)
5cos%5E2%28x%29%28cos%28x%29+-+4%29+%2B+1%28cos%28x%29+-+4%29+=+0
As we can see we have a common factor between the two groups: cos(x) - 4. We can factor out this common factor:
%285cos%5E2%28x%29+%2B+1%29%28cos%28x%29+-+4%29+=+0
From the Zero Product Property:
5cos%5E2%28x%29+%2B+1+=+0 or cos%28x%29+-+4+=+0
Solving the first one...
5cos%5E2%28x%29+=+-1
cos%5E2%28x%29+=+-1%2F5
At this point we should realize that there is no solution to this equation. Squaring a cos (or anything for that matter) cannot result in a negative number.

Solving the second equation:
cos%28x%29+-+4+=+0
cos%28x%29+=+4
Again we should recognize that there is no solution. A cos cannot be larger than 1!

So your equation has no solutions!