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Solve the equation $$\cos x -2\cos 2x+3 \cos 3x -4\cos 4x = \dfrac{1}{2}.$$ I tried, put $t = \cos x$.

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    Yes, and what happened when you did that?2012-11-04
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    Did you try the angle sum identity to break up some of those cosines? Remember: $cos(2x)=cos^2(x)-sin^2(x)$.2012-11-04
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    Or better still $\cos 2x = 2 \cos^2 x - 1$. It should be possible to turn this into a big polynomial in $\cos x$.2012-11-04
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    Note that $$ \begin{align} \sum\limits_{k=1}^n (-1)^k\sin kx&=\frac{1}{2}\sec\frac{x}{2}\sum\limits_{k=1}^n (-1)^k 2\sin kx\cos\frac{x}{2}\\ &=\frac{1}{2}\sec\frac{x}{2}\sum\limits_{k=1}^n (-1)^k \left(\sin\left(k+\frac{1}{2}\right)x+\sin\left(k-\frac{1}{2}\right)x\right)\\ &=\frac{1}{2}\sec\frac{x}{2}\left(-\sin\frac{x}{2}+(-1)^n\sin\left(n+\frac{1}{2}\right)x\right)\\ \end{align} $$2012-11-04
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    hence $$ \begin{align} \sum\limits_{k=1}^n(-1)^{k-1}k\cos kx &=-\frac{d}{dx}\left(\sum\limits_{k=1}^n (-1)^k\sin kx\right)\\ &=-\frac{d}{dx}\left(\frac{1}{2}\sec\frac{x}{2}\left(-\sin\frac{x}{2}+(-1)^n\sin\left(n+\frac{1}{2}\right)x\right)\right)\\ &=\frac{1}{4}\sec^2\frac{x}{2}(1-(n+1)(-1)^n\cos nx-n(-1)^n\cos(n+1)x) \end{align} $$ Setting $n=4$ we get $$ \cos x-2\cos 2x+3\cos 3x-4\cos 4x=\frac{1}{4}\sec^2\frac{x}{2}(1-5\cos 4x-4\cos 5x) $$2012-11-04
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    Thus we get an equation $$ \frac{1}{4}\sec^2\frac{x}{2}(1-5\cos 4x-4\cos 5x)=\frac{1}{2} $$ $$ 5\cos 4x+4\cos 5x+\cos x=0 $$ But I don't know whether this is a right way2012-11-04
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    I think you want to have something factored equal to 0, and then set each factor to 0. But you have two factors multiplying to give 1/2 here.2012-11-04

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With $c=\cos x$ we have $\cos 2x=2c^2-1$, and $\cos 3x = 4c^3-3c$, and finally $\cos 4x = 8c^4-8c^2+1$. If you take your equation, move the 1/2 to the left side, make the above substitutions, and multiply by $-2$, you'll get $$64c^4-24c^3-56c^2+16c+5=0.$$ This factors as $(2c-1)(32c^3+4c^2-26c-5)=0$. The cubic here has no rational roots, and it looks like one would have to resort to the cubic equation (a mess) to find its zeros. Numerically the zeros of the cubic are about $-.859,-.195,+.929$, all can be cosine of an angle. So from the other root $c=1/2$ of the linear factor, you'll have a total of eight solutions in each interval $[2k \pi,(2k+2)\pi]$. Looks like only the solutions from $ \cos x=1/2$ will be familiar angles: $\pi/3$ and $5\pi/3$ and their translates by $2 \pi n$.