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 816780: 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)(8cos(x) + 4) = 3
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
The left side is factored. But the right side is not zero. So having the left side factored does not help. Multiplying out the left side:
Now we get a zero on the right by subtracting 3:
Now we factor, if we can. But this will not factor. It is a quadratic however. So we can use the quadratic formula with an "a" of 8, a "b" of 4 and a "c" of -3:
Simplifying...
which is short for:
or
Using our calculator to get decimal approximations (and rounding them to 4 places):
or
Simplifying:
or
Using the inverse cos on our calculator on these decimals...
For , is 1.1467. So the reference angle is 1.1467. And since the 0.4115 is positive and since cos is positive in the 1st and 4th quadrants we should get general solution equations of:
(for the 1st quadrant)
(for the 4th quadrant)
For , is 0.4239. (Note: Do not enter the minus of -0.9115 when looking for a reference angle! With a minus on the decimal, , we will get second quadrant angle, not a reference angle!) So the reference angle is 0.4239. And since the -0.9115 is negative (here is where the minus gets used) and since cos is negative in the 2nd and 3rd quadrants we should get general solution equations of:
(for the 2nd quadrant)
(for the 3rd quadrant)
Replacing first 's with 3.1416:
(for the 2nd quadrant)
(for the 3rd quadrant)
which simplifies to:
So the general solution equations are:
From these we will try various integer values for n looking for x's that are in the given interval.
From ...
if n = 0 then x = 1.1467
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 ...
if n = 0 (or smaller) then x is too small for the interval
if n = 1 then x = 1.1467 + = 5.1365
if n = 2 (or larger) then x is too large for the interval
From ...
if n = 0 then x = 2.7177
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 ...
if n = 0 then x = 3.5655
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 specific solutions within the given interval are:
1.1467, 5.1365, 2.7177 and 3.5655
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