We'll write (sin x)^2 with respect to (cos x)^2, from the
fundamental formula of trigonomtery:
(sin x)^2 = 1 - (cos
x)^2 (1)
We'll substitute (1) in the given
equation:
cos x = 1 + 1 - (cos
x)^2
We'll combine like terms and we'll move all terms to
the right side:
(cos x)^2 + cos x - 2 =
0
We'll substitute cos x by
t:
t^2 + t - 2 = 0
We'll apply
the quadratic formula:
t1 = [-1+sqrt(1 +
8)]/2
t1 = (-1+3)/2
t1 =
1
t2 = (-1-3)/2
t2 =
-2
But t1 = cos x => cos x =
1
x = +/-arccos (1) +
2k*pi
x = 0 +
2k*pi
x =
2kpi
We'll reject the second
solution for t since -1 =< cos x =<
1
No comments:
Post a Comment