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[Nevermore]

Oh, you got a hard nut there:P

Anyway, that's my solution. Basic idea: Use a sum formula for cosinus, use the identity sin^2 + cos^2 = 1, substitute sin(x), solve the resulting quadratic equation and resubstitute.

First off, I don't like to type Theta all the time, so I just use x.

That would make:

3cos(2x) + sin(x)-1 = 0 <=>

3cos(x+x) + sin(x) = 1 <=>

3(cos^2(x)-sin^2(x)) + sin(x) = 1 <=> //Used the sum formula

3((1-sin^2(x))-sin^2(x)) + sin(x) = 1 <=> //Identity cos^2 = 1-sin^2(x)

3 - 6sin^2(x) + sin(x) = 1 <=>

sin^2(x) - (1/6)sin(x) = 1/3 <=>

u^2 - (1/6)u = 1/3 with substitution u=sin(x) <=>

(u - (1/12))^2 = 49/144 <=> //quadratic complement

|u - (1/12)| = 7/12 <=>

u1 = -1/2, u2 = 2/3

Resubstitution:

u1 = sin(x1) <=> x1 = arcsin(-1/2) = pi/6
u2 = sin(x2) <=> x2 = arcsin(2/3) = 0,7297276562...

Edit: Ah, I forget the other solutions... So, everything in the form of 2pi*k + x1 or 2pi*k + x2 are solutions, with k being an integer.

---

But the Raven, sitting lonely on that placid bust, spoke only
That one word, as if his soul in that one word he did outpour.
Nothing further then he uttered; not a feather then he fluttered--
Till I scarcely more than muttered, &quot;Other friends have flown before--
On the morrow he will leave me, as my Hopes have flown before.&quot;
<blockquote>Then the bird said, &quot;Nevermore.&quot;</blockquote>
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