SOLUTION: cos^2x-5cosx+4=0

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Question 812608: cos^2x-5cosx+4=0
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
%28cos%28x%29%29%5E2-5cos%28x%29%2B4 could be called a quadratic polynomial in cos%28x%29.
It has a term in cos%28x%29 squared, a term in cos%28x%29, and an independent term.
It can be factored as
%28cos%28x%29%29%5E2-5cos%28x%29%2B4=%28cos%28x%29-1%29%28cos%28x%29-4%29 ,
which means the equation can be re-written as
%28cos%28x%29-1%29%28cos%28x%29-4%29=0 .
That is not easy to see, but someone saw something like that before,
and invented a trick called "change of variable".

If you make the change of variable y=cos%28x%29 ,
substituting y for cos%28x%29 , you get
y%5E2-5y%2B4=0

That equation is not that scary,
and you probably would solve it easily by factoring,
first re-writing it as the factored form
%28y-1%29%28y-4%29=0 ,
and then finding that the solutions derive from
y-1=0 --> highlight%28y=1%29
and
y-4=0 --> highlight%28y=4%29

That would mean (going back to the original variable)
cos%28x%29=1 , which has many solutions,
and
cos%28x%29=4 , which has no solutions.
cos%280%29=1 , so cos%28x%29=1
works for the angle measuring 0%5Eo in degrees,
which is also known as 0 radians (where radian is a ratio of length with no units).
Since angles are useful to measure turns (of a screw, or a knob, or a skater),
we can talk about angles measuring 360%5Eo, or 720%5Eo
(that is 2pi%7D%7D+radians+or+%7B%7B%7B4piradians),
or greater multiples.
When we care about the direction of the turn,
turns in one direction are counted as positive angles,
and turns in the other direction count as negative angles,
so there is an infinite number of solutions.
We could express them all with a formula.
In degrees, it would be
highlight%28k%2A360%5Eo%29 where k is any integer.
In radians, it would be
highlight%282k%2Api%29 where k is any integer.