We'll write (cos x)^2 with respect to (sin x)^2, from the
fundamental formula of trigonomtery:
(cos x)^2 = 1 - (sin
x)^2 (1)
We'll substitute (1) in the given
equation:
sin x = 1 + 1 - (sin
x)^2
We'll combine like terms and we'll move all terms to
the right side:
(sin x)^2 + sin x - 2
=0
We'll substitute sin 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 = sin x => sin x =
1
x = (-1)^k*arcsin (1) +
k*pi
x = (-1)^k*(pi/2)
+ k*pi
We'll reject the
second solution for t since -1 =< sin x =<
1.
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