Expert: Scotto - 2/10/2009

Question
Use factoring, the quadratic formula, or identities to solve the equation. Find all solutions in the interval [0, 2π) and list them in increasing order.



5cos^2x + 17cosx = 12



(2 possible answers)

x=

x=


When I factored these out I got 3/5 and -4. However, neither work..when I try putting in cos^-1 of (3/5 and -4) please help! thanks.

Answer
If you let a = cos(x), there is the quadratic 5a² + 17a - 12 = 0.

This, as you stated, factors into (5a-3)(a+4), which makes
a be either 3/5 or -4.  NOte that a is cos(x), and it is
known that |cos(x)|≤1, so even though a could be -4, when
it is looked into the fact that cos(x) = a, this is not
possible.

However, for cos(x) = 3/5, that would mean if we took a triangle
with the near side 3 and the hypoteneuse 5, the far side would
be √(5²-3²) = √(25-9) = √16 = ±4.

Since the near side is positive, it is in quadrant I or IV.
I get 0.674740942 radians or 38.65980825°.

Note that if these two value were subtracted from 0, they would
also work.  However, to be in the interval [0,2π], you have to
add 2π to them (or 360°).

When I first started working with a calculator the could do trig
functions, occasionally I put in 3.14 (which is close to pi) and
computed the sin() to make sure I was in radian format.  It didn't
tell me, it just had one little dot and it took me awhile to learn
what that one little dot meant.  Boy, that was along time ago - back in the 70s.  Did you know that 1976 was the 200th anniversary of our
country (Yeah, you know, we revolted from England in 1776)?  There
were a lot of fireworks that year.  It was also the year that I
graduated from 8th grade, my sister graduated from high school,
and one of my brothers (I have 2) graduated from college.
All together, it was a pretty special year to the country
and to my family.