Question

The general solution of $\cos 3x+8{\cos }^{3}x=0$ is .

• A. $x\in \left\{\right(2n+1\left)\frac{\pi }{2}\right\}\cup \{n\pi +\frac{\pi }{3}\},n\in Z$
• B. $x\in \left\{\right(2n+1\left)\frac{\pi }{2}\right\}\cup \{n\pi \pm \frac{\pi }{3}\},n\in Z$
• C. $\mathrm{Not}\mathrm{possible}\mathrm{to}\mathrm{determine}.$

Answer 1

The correct option is B $x\in \left\{\right(2n+1\left)\frac{\pi }{2}\right\}\cup \{n\pi \pm \frac{\pi }{3}\},n\in Z$

Given: $\cos 3x+8{\cos }^{3}x=0$

To find the general solution, follow the steps:-

1. Transform the equation into product of linear factors
2. Equate each factor to zero
3. Solve each equation individually
4. Take union of all these solutions

To make the arguments of cosine uniform, we apply the formula $\cos 3x=4{\cos }^{3}x-3\cos x$

The resulting equation is
$4{\cos }^{3}x-3\cos x+8{\cos }^{3}x=0$
$\Rightarrow 12{\cos }^{3}x-3\cos x=0$
(Dividing by 3 on both side)
$\Rightarrow 4{\cos }^{3}x-\cos x=0$
$\Rightarrow \cos x(4{\cos }^{2}x-1)=0$
$\Rightarrow (\cos x-0)(4{\cos }^{2}x-1)=0$

Here, either $\cos x=0$ or $4{\cos }^{2}x-1=0.$

For $\cos x=0,$ general solution is $x=(2n+1)\frac{\pi }{2},n\in Z.$
Represent the above solution set as $A.$

For $4{\cos }^{2}x-1=0,$
${\cos }^{2}x=\frac{1}{4}$
$\Rightarrow {\cos }^{2}x=(\frac{1}{2}{)}^2$
$\Rightarrow {\cos }^{2}x=(\cos \frac{\pi }{3}{)}^2$
The general solutoin, say B, is $x=n\pi \pm \frac{\pi }{3},n\in Z.$

The complete general solution is $A\cup B,$ i.e., $x\in \{(2n+1)\frac{\pi }{2}\}\bigcup \{n\pi \pm \frac{\pi }{3}\},n\in Z$