2
$\begingroup$

I want to solve $$\cos 3x - 18\cos x +10 =0 $$

I tried:

1) Replacing $\cos 3x$ to $\cos^3x - 3\cos x$

2) Replacing $\cos x$ to $t$ we get:

$$t^3 - 21t +10 = 0$$

So we get cubic equation. But I can't solve it.

$\endgroup$

2 Answers 2

7
$\begingroup$

You are wrong. $\cos 3x = 4\cos^3 x-3\cos x$. Denote $t=\cos x$; we have $$ 4t^3-21t+10=0\Longrightarrow t = -\frac52,2,\frac12. $$ $t=-5/2$ and $t=2$ are not suitable, of course.

$\endgroup$
4
$\begingroup$

Given $$cos3x - 18cosx +10 =0 $$ $$\implies 4\cos^3x-3\cos x-12\cos x+10 = 0$$ Factorizing the cubic equation as follows $$\left(\cos x-\frac{1}{2}\right) \left(\cos x-2\right)\left(\cos x+\frac{5}{2}\right)= 0$$ $$\text{if}\ \cos x-\frac{1}{2}=0 \implies \cos x=\frac{1}{2}\iff x=2n\pi\pm\frac{\pi}{3} $$ $$\text{if}\ \cos x-2=0 \implies \cos x\neq 2$$$$\text{if}\ \cos x+\frac{5}{2}=0 \implies \cos x\neq \frac{-5}{2}$$ Hence, the general solution for $x$ is $$x=2n\pi\pm \frac{\pi}{3}$$ Where, $n$ is any integer

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .