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solve for x:$$\cos^2 3x+\frac{\cos^2 x}{4}=\cos 3x\cos^4 x$$

My attempt:

completing square: $${(\cos 3x-\frac{\cos x}{2})}^2=\cos 3x\cos x (1-\cos^3 x)$$ or $$\cos x\cos 3x\ge 0$$ also by some basic identities :$$1+\cos 6x+\frac{1+\cos 2x}{4}=(\cos 4x+\cos 2x)\left(\frac{\cos 3x+3\cos x}{4}\right)$$

I cant proceed now.....

Albus Dumbledore
  • 11,860

1 Answers1

2

$$\color{blue}{\cos^2{3x}} - (\cos^4{x})\color{blue}{\cos{3x}} + \dfrac{\cos^2{x}}{4}=0$$

$$ \color{blue}{\cos{3x}} = \dfrac{\cos^4{x} \pm \sqrt{\cos^8{x} - \cos^2{x}}}{2}$$

Use $\cos^8{x} - \cos^2{x} \le 0$.

Similarly,

$$\cos{3x}\color{red}{\cos^4{x}} - \dfrac{\color{red}{\cos^2{x}}}{4} - \cos^2{3x}=0$$

cosmo5
  • 10,629