Question

General solution of the equation
$3\sqrt{3}si{n}^{3}x+co{s}^{3}x+3\sqrt{3}sinxcosx=1$ is

• A. $n\pi +(-1{)}^n\frac{\pi }{6};n\in l$
• B. $2n\pi ,n\in l$
• C. $n\pi +(-1{)}^n\frac{\pi }{6}-\frac{\pi }{6}$
• D. none of these

Answer 1

The correct option is C $n\pi +(-1{)}^n\frac{\pi }{6}-\frac{\pi }{6}$

$3\sqrt{3}si{n}^{3}x+co{s}^{3}x+3\sqrt{3}sinxcosx=1$
$3\sqrt{3}si{n}^{3}x+co{s}^{3}x-1=3\left(\sqrt{3}sinx\right)\left(cosx\right)\left(1\right)$
We know that
If $a+b+c=0\Rightarrow a^3+b^3+c^3=3abc$
Now $a=\sqrt{3}sinx,b=cosx,c=-1$
$a^3+b^3+c^3-3abc=0$
Possible only if $a+b+c=0$
$\sqrt{3}sinx+cosx-1=0\Rightarrow \sqrt{3}sinx+cosx=1$
$\frac{\sqrt{3}}{2}sinx+\frac{1}{2}cosx=\frac{1}{2}\Rightarrow sin\left(\begin{array}{c}x+\frac{\pi }{6}\end{array}\right)=sin\frac{\pi }{6}$
$\Rightarrow x+\frac{\pi }{6}=n\pi +(-1{)}^n\pi /6\Rightarrow x=n\pi +(-1{)}^n\frac{\pi }{6}-\frac{\pi }{6}$