Question 816780
{{{cos(x)(8cos(x) + 4) = 3}}}
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:
{{{8cos^2(x) + 4cos(x) = 3}}}
Now we get a zero on the right by subtracting 3:
{{{8cos^2(x) + 4cos(x) - 3=0}}}
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:
{{{cos(x) = (-(4)+-sqrt((4)^2-4(8)(-3)))/2(8)}}}
Simplifying...
{{{cos(x) = (-4+-sqrt(16-4(8)(-3)))/2(8)}}}
{{{cos(x) = (-4+-sqrt(16+96))/16}}}
{{{cos(x) = (-4+-sqrt(112))/16}}}
{{{cos(x) = (-4+-sqrt(4*28))/16}}}
{{{cos(x) = (-4+-sqrt(4*4*7))/16}}}
{{{cos(x) = (-4+-sqrt(4)*sqrt(4)*sqrt(7))/16}}}
{{{cos(x) = (-4+-2*2*sqrt(7))/16}}}
{{{cos(x) = (-4+-4sqrt(7))/16}}}
{{{cos(x) = (4(-1+-sqrt(7)))/16}}}
{{{cos(x) = (-1+-sqrt(7))/4}}}
which is short for:
{{{cos(x) = (-1+sqrt(7))/4}}} or {{{cos(x) = (-1-sqrt(7))/4}}}<br>
Using our calculator to get decimal approximations (and rounding them to 4 places):
{{{cos(x) = (-1+2.6458)/4}}} or {{{cos(x) = (-1-2.6458)/4}}}
Simplifying:
{{{cos(x) = 0.4115}}} or {{{cos(x) = -0.9115}}}<br>
Using the inverse cos on our calculator on these decimals... 
For {{{cos(x) = 0.4115}}}, {{{cos^-1(0.4115)}}} 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:
{{{x = 1.1467 + 2pi*n}}} (for the 1st quadrant)
{{{x = -1.1467 + 2pi*n}}} (for the 4th quadrant)<br>
For {{{cos(x) = -0.9115}}}, {{{cos^-1(0.9115)}}} is 0.4239. (Note: Do not enter the minus of -0.9115 when looking for a reference angle! With a minus on the decimal, {{{cos^-1(0.9115)}}}, we will get second quadrant angle, not a reference angle!) So the reference angle is 0.4239. And since the -0.9115 is negative (<i>here</i> is where the minus gets used) and since cos is negative in the 2nd and 3rd quadrants we should get general solution equations of:
{{{x = pi+0.4239 + 2pi*n}}} (for the 2nd quadrant)
{{{x = pi-0.4239 + 2pi*n}}} (for the 3rd quadrant)
Replacing first {{{pi}}}'s with 3.1416:
{{{x = 3.1416-0.4239 + 2pi*n}}} (for the 2nd quadrant)
{{{x = 3.1416+0.4239 + 2pi*n}}} (for the 3rd quadrant)
which simplifies to:
{{{x = 2.7177 + 2pi*n}}}
{{{x = 3.5655 + 2pi*n}}}<br>
So the general solution equations are:
{{{x = 1.1467 + 2pi*n}}}
{{{x = -1.1467 + 2pi*n}}}
{{{x = 2.7177 + 2pi*n}}}
{{{x = 3.5655 + 2pi*n}}}
From these we will try various integer values for n looking for x's that are in the given interval.
From {{{x = 1.1467 + 2pi*n}}}...
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 {{{x = -1.1467 + 2pi*n}}}...
if n = 0 (or smaller) then x is too small for the interval
if n = 1 then x = 1.1467 + {{{2pi}}} = 5.1365
if n = 2 (or larger) then x is too large for the interval
From {{{x = 2.7177 + 2pi*n}}}...
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 {{{x = 3.5655 + 2pi*n}}}...
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<br>
So the specific solutions within the given interval are:
1.1467, 5.1365, 2.7177 and 3.5655