SOLUTION: Use inverse trigonometric functions to find the solutions of the equation that are in the interval [0, 2π).You may enter an exact answer or round your solutions to four decima

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Question 816469: Use inverse trigonometric functions to find the solutions of the equation that are in the interval [0, 2π).You may enter an exact answer or round your solutions to four decimal places (this is best for this problem). (Enter your 4 answers in a comma separated list.)
cos(x)(15cos(x) + 7) = 3
Found 2 solutions by lwsshak3, jsmallt9:
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
cos(x)(15cos(x) + 7) = 3
15cos^2(x)=3-7=-4
cos^2(x)=-4/15
no solution:
cos^2(x)≥0

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
There's an error in the other tutor's response so his answer is not right. When you multiply out
cos(x)(15cos(x) + 7) = 3
you get
15cos%5E2%28x%29%2B7cos%28x%29=3
not
15cos%5E2%28x%29%2B7=3

So the equation to solve is:
15cos%5E2%28x%29%2B7cos%28x%29=3
To solve this we subtract 3:
15cos%5E2%28x%29%2B7cos%28x%29-3=0
This equation is in what is called quadratic form. This is an equation with the same structure as a "pure" quadratic and can be solved using the same techniques. To help you see the "quadratic-ness" of this equation we'll use a temporary variable:
Let q = cos(x). Replacing the cos(x) with q in the equation we get:
15q%5E2%2B7q-3=0
This is clearly a quadratic. It does not factor but we can use the quadratic formula:
q+=+%28-%287%29%2B-sqrt%28%287%29%5E2-4%2815%29%28-3%29%29%29%2F2%2815%29
Simplifying...
q+=+%28-7%2B-sqrt%2849-4%2815%29%28-3%29%29%29%2F30
q+=+%28-7%2B-sqrt%2849%2B180%29%29%2F30
q+=+%28-7%2B-sqrt%28229%29%29%2F30
which is short for
q+=+%28-7%2Bsqrt%28229%29%29%2F30 or q+=+%28-7-sqrt%28229%29%29%2F30

Of course we do not care about values for q. We want solutions for x. So we replace the q's...
cos%28x%29+=+%28-7%2Bsqrt%28229%29%29%2F30 or cos%28x%29+=+%28-7-sqrt%28229%29%29%2F30
... and then solve for x. We should know that neither of the right sides of these equations are special angle values for cos. So we reach for our calculators to convert the right sides into decimals (rounded-off to 4 places):
cos(x) = 0.2711 or cos(x) = -0.7378

Now we use our calculators again to find reference angles. cos%5E-1%280.2711%29+=+1.2963. Since the 0.2711 is positive and since cos is positive in the 1st and 4th quadrants we should get the following general solution equations for cos(x) = 0.2711:
x+=+1.2963+%2B+2pi%2An (for the 1st quadrant)
x+=+-1.2963+%2B+2pi%2An (for the 4th quadrant)

For cos(x) = -0.7378, cos%5E-1%280.7378%29+=+0.7410 (Note: Do not use a minus sign when finding reference angles!). Since the -0.7378 is negative (this is where the negative gets used) and since cos is negative in the 2nd and 3rd quadrants we should get the following general solution equations:
x+=+pi+-+0.7410+%2B+2pi%2An (for the 2nd quadrant)
x+=+pi+%2B+0.7410+%2B+2pi%2An (for the 3rd quadrant)
Using 3.1316 for pi these simplify to:
x+=+2.4006+%2B+2pi%2An
x+=+3.8826+%2B+2pi%2An

So the general solution equations for your equation are:
x+=+1.2963+%2B+2pi%2An
x+=+-1.2963+%2B+2pi%2An
x+=+2.4006+%2B+2pi%2An
x+=+3.8826+%2B+2pi%2An

Now we use these equations and various integer values for n to find x's that are in the specified domain.
From x+=+1.2963+%2B+2pi%2An...
if n = 0 then x = 1.2963
if n = 1 (or larger) then x is too large for the interval
if n = -1 (or smaller) then x is too small for the interval
From x+=+-1.2963+%2B+2pi%2An...
if n = 0 (or smaller) then x is too small for the interval
if n = 1 then x = 1.8453
if n = 2 (or larger) then x is too large for the interval
From x+=+2.4006+%2B+2pi%2An...
if n = 0 then x = 2.4006
if n = 1 (or larger) then x is too large for the interval
if n = -1 (or smaller) then x is too small for the interval
From x+=+3.8826+%2B+2pi%2An...
if n = 0 then x = 3.8826
if n = 1 (or larger) then x is too large for the interval
if n = -1 (or smaller) then x is too small for the interval

So the only solutions within the given interval are 1.2963, 1.8453, 2.4006 and 3.8826

P.S. After you've done a few of these quadratic form equations you will not need the temporary variable. You will start seeing how to go directly from
15cos%5E2%28x%29%2B7cos%28x%29-3=0
to
cos%28x%29+=+%28-%287%29%2B-sqrt%28%287%29%5E2-4%2815%29%28-3%29%29%29%2F2%2815%29
etc.