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We have to find all value of $\theta,$ between $0$ & $\pi,$ which satisfy the equation; $\cos\theta \cdot \cos2\theta \cdot \cos 3\theta = \dfrac{1}{4}$

I tried it a lot.

But every time I got stuck and get no result.

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  • $\begingroup$ Use $\cos(A)\cos(B) = \dfrac{1}{2}\times(\cos(A+B)+\cos(A-B))$ $\endgroup$ – Max Payne Dec 28 '16 at 6:28
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We have, $$\cos \theta \cos 2\theta \cos 3\theta =\frac{1}{4}$$ $$\Rightarrow 4\cos \theta \cos 2\theta \cos 3\theta =1$$ $$\Rightarrow (2\cos 3\theta \cos \theta)2\cos 2\theta =1$$ $$\Rightarrow (\cos 4\theta +\cos 2\theta)2\cos 2\theta -1=0$$ $$\Rightarrow 2\cos 2\theta \cos 4\theta +\cos 4\theta =0$$ This gives us, $$\cos 4\theta =0\mid \cos 2\theta =-\frac{1}{2}$$ where $\mid$ stands for "or". I hope you will be able to take it from here.

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HINT:

Use Werner Formulas $2\cos x\cos3x=\cos2x+\cos4x$

and $2\cos4x=\cos^22x-1$ to form a cubic equation in $\cos2x$

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  • $\begingroup$ But how to solve cubic equation $\endgroup$ – user123733 Dec 28 '16 at 6:38
  • $\begingroup$ $$4c^3+2c^2-2c-1=(2c+1)2c^2-(2c+1)=?$$ $\endgroup$ – lab bhattacharjee Dec 28 '16 at 6:46

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