Solve 8 cos2 (1 - x ) + 13 cos(1 - x )- 5 = 0 .
Solution
Now, as specified prior to starting the instance this quadratic does not factor. Though, that doesn't mean all is lost. We can solve out the following equation along with the quadratic formula (you do remember this & how to employ it right?),
8t 2 + 13t - 5 = 0 ⇒ t= (-13 ±√329)/ 16= 0.3211,-1.9461
Hence, if we can employ the quadratic formula on this then we can also employ it on the equation we're asked to solve. Doing this gives us,
Cos(1 - x ) = 0.3211 OR cos(1 - x ) = -1.9461
Now, recall previous section. In that example we noted that
-1 ≤ cos(θ ) ≤ 1 and hence the second equation will have no solutions. Thus, the solutions to the first equation will yield the only solutions to our original equation. Solving out this gives the given set of solutions,
x= -0.2439 - 2 ∏ n
x= -4.0393 - 2 ∏ n
n= 0, ±1, ±2,..........
Note as well that we did get some negative numbers here and that does appear to violate the general form which we've been using in most of these examples. Though, in this case the "-" are coming about while we solved for x after calculating the inverse cosine in our calculator.
There is one example more in this section that we have to work that illustrates another way wherein factoring can arise in solving trig equations. This equation is also the only one where the variable seems both inside & outside of the trig equation. Not all equations in this form can be solved easily; though some can so we desire to do a quick example of one.