Analyze and solve $m \cos^3 x - \sin 2x \sin x-\cos x =0$

Question

Could you help me with this equation?

$m\cos^3 x-\sin2x \sin x-\cos x=0$

I can solve this one with $m=1$, which is not hard. It seems to me that I need to convert $\sin 2x \sin x$ into some forms of cosine and eventually factorize it, but I can't seem to find a way.

Thank you!

Answer 1

\begin{align} & m \cos^3 (x) - \sin(2x) \sin (x)- \cos (x) =0 \\ \implies & m \cos^3 (x) - 2\sin(x)\cos(x)\sin(x)-\cos(x)=0 && \text{double angle formula}\\ \implies & m \cos^3 (x) - 2\cos(x)\sin ^2(x)-\cos(x)=0 && \text{grouping the sines}\\ \implies & m \cos^3 (x) - 2\cos(x)\bigl (1-\cos ^2 (x) \bigr)-\cos(x)=0 && \sin^2(x) = 1-\cos^2(x)\\ \implies & (m+2) \cos^3 (x) - 3\cos(x)=0 && \text{group like terms}\\ \implies & \cos^3 (x) - \frac{3}{m+2}\cos(x)=0 && \text{assume $m \neq -2$}\\ \implies & \cos(x) \biggl [\biggl (\cos(x)+\sqrt{\frac{3}{m+2}} \biggr)\biggl (\cos(x)-\sqrt{\frac{3}{m+2}} \biggr) \biggl]=0 && \text{factorizing}\\ \implies & \cos(x) = 0 \quad \cos(x)=\pm\sqrt{\frac{3}{m+2}} \end{align}

$\cos(x) = 0 \implies x = (2n+1) \pi$ for any $n \in \Bbb Z$

$\cos(x)=\pm\sqrt{\frac{3}{m+2}}$ has real solutions only when $m≥1$.

Answer 2

$$ m\cos^{3}x-\sin {2x}\sin x-\cos x=0$$

$$ m\cos^{3}x-2\sin^ {2}x \cos x-\cos x=0$$

One solution comes from $ \cos x =0$

$$ m\cos^{2}x-2 (1- \cos^ {2}x) x-1=0$$

$$ \cos ^2 x = \frac {3}{m+2}$$

Answer 3

Thanks, for your fast reply, the solution is like this though: $\implies m\cos^3x-\frac{1}{2}(\cos x-\cos3x)-\cos x=0$

$\implies (m-\frac{1}{2})\cos^3x-\frac{1}{2}\cos x=0$ $[1]$

$\implies (m-\frac{1}{2})(4\cos^3x-3\cos x)-\frac{1}{2}\cos x=0$ $\implies \cos x[(4m-2)\cos^2x-3m+1]=0$

$\implies \cos x=0 => x=\frac{\pi}{2}+2k\pi$ (1)

$\implies (4m-2)\cos^2x-3m+1=0$ (2)

Analyse (2), $m=\frac{1}{2}$ the equation has no roots

$m$ is not $\frac{1}{2}$, then $\cos^2x=\frac{3m-1}{4m-2}$

$\cos x=\pm\sqrt{\frac{3m-1}{4m-2}}$ only when $-1<\frac{3m-1}{4m-2}<1$

Do you think the solution is correct?

Edit: The derivation $sin2xsinx=\frac{1}{2}(cosx-cos3x)$is as followed: $sin2xsinx = (2sinxcosx)sinx=2sin^2xcosx=(1-cos2x)cosx=(2-2cos^2x)cosx=2cosx-2cos^3x=\frac{1}{2}cosx-\frac{4}{3}cos^3x+\frac{3}{2}cosx=\frac{1}{2}(cosx-cos3x)$

This is quite long, if you ask me.