What is x if cos^2(x/2)-2cos^2x=(3/2)*square root2(1+cosx)+2sin^2x?
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We have to solve cos^2(x/2)-2cos^2x=(3/2)*square root2(1+cosx)+2sin^2x for x.
Rewriting the equation in a simpler form:
(cos x/2)^2 - 2 ( cos x)^2 = (3/2)sqrt 2 ( 1 + cos x) + 2 (sin x)^2
we know that...
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The first step is to move all terms to the left side side:
(cos x/2)^2 - 2(cos x)^2 - (3/2)*square root2(1+cosx) - 2(sin x)^2 = 0
We recognize the formula for the half angle:
square root[(1+cosx)/2] = (cos x/2)
We'll re-group the terms:
(cos x/2)^2 - 2[(sin x)^2 + (cos x)^2] - [3*2(cos x/2)]/2= 0
We'll write the fundamental formula of trigonometry;
[(sin x)^2 + (cos x)^2] = 1
The equation will become:
(cos x/2)^2 - 2- 3(cos x/2) = 0
We'll substitute cos x/2 = t
t^2 - 3t - 2 = 0
We'll apply quadratic formula:
t1 = [-3 + sqrt(9 + 8)]/2
t1 = (-3 + sqrt17)/2
t2 = (-3 - sqrt17)/2
But cos x/2 = t
cos x/2 = t1
cos x/2 = (-3 + sqrt17)/2
x = +/-2arccos[(-3 + sqrt17)/2] + 2kpi
cos x/2 = t2
cos x/2 = (-3 - sqrt17)/2, impossible since cos a > = - 1.
The only solution of the equation is: {+/-2arccos[(-3 + sqrt17)/2] + 2kpi}.
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