To solve the equation, we'll try to write it using just a
single function.
We'll write cos 2x = cos (x+x) = cosx*cosx
- sinx*sinx = (cos x)^2 - (sin x)^2
We'll substitute (cos
x)^2 = 1 - (sin x)^2 in the formula of cos 2x:
cos 2x = 1 -
(sin x)^2 - (sin x)^2
cos 2x = 1 - 2(sin
x)^2
The equation will
become:
1 - 2(sin x)^2 + sin x =
0
- 2(sin x)^2 + sin x + 1 =
0
We'll multiply by -1:
2(sin
x)^2 - sin x - 1 = 0
We'll substitute sin x =
t
2t^2 - t - 1 = 0
We'll apply
the quadratic formula:
t1 =
[1+sqrt(1+8)]/4
t1 =
(1+3)/4
t1 = 1
t2 =
-2/4
t2 = -1/2
So, sin x =
t1
sin x = 1
x = arcsin (1) +
k*pi
x = pi/2 + k*pi
sin x =
-1/2
x = (-1)^(k+1) * arcsin (1/2) +
k*pi
x = (-1)^(k+1) *(pi/6) +
k*pi
k is an integer number.
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