How do you solve #cos^2x+2cosx+1=0# over the interval 0 to 2pi?
1 Answer
Apr 2, 2016
Solve as a quadratic first to find the value of
Explanation:
Factorize the left hand side.
#cos^2(x) + 2cos(x) + 1 = (1 + cos(x))^2 = 0#
This means that
#1 + cos(x) = 0#
or
#cos(x) = -1#
From the graph of
graph{cos(x) [-10, 10, -5, 5]}
The only value of