Question 1002887
Find the exact value of
sin (x/2) = 2cos^2(x)-1
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The trig functions have to have the same argument.
sin^2(x/2) = (1 - cos(x))/2
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sin(x/2) = 2cos^2(x)-1
Square both sides
sin^2(x/2) = 4cos^4(x) - 4cos^2(x) + 1
(1 - cos(x))/2 = 4cos^4(x) - 4cos^2(x) + 1
1 - cos(x) = 8cos^4(x) - 8cos^2(x) + 2
8cos^4(x) - 8cos^2(x) + cos(x) + 1 = 0
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If you inspect
8x^4 - 8x^2 + x + 1 = 0 it's obvious that x = -1 is a zero.
--> cos(x) = -1
--> x = pi + n*2pi, n = ±0,1,2,3...
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Divide the quartic by (x+1)
--> 8x^3 - 8x^2 + 1 = 0
It's less obvious, x = 1/2 is a zero.
--> cos(x) = 0.5
--> x = pi/3 + n*2pi, n = ±0,1,2,3...
and
--> x = 2pi/3 + n*2pi, n = ±0,1,2,3...
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Divide the cubic by (x - 1/2):
--> 8x^2 - 4x - 2 = 0
*[invoke solve_quadratic_equation 8,-4,-2]

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--> cos(x) = {{{1/4 +-sqrt(5)/4}}}
--> x = pi/5 + n*2pi, n = ±0,1,2,3...
--> x = 9pi/5 + n*2pi, n = ±0,1,2,3...
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and x = 3pi/5 + n*2pi, n = ±0,1,2,3...
and x = 7pi/5 + n*2pi, n = ±0,1,2,3...