Question

${\sin }^{3}x+{\sin }^3({120}^0+x)+{\sin }^3({240}^0+x)\in$ $[-\frac{p}{4},\frac{p}{4}]\supset p=$

• A. $1$
• B. $-1$
• C. $4$
• D. $3$

Answer 1

The correct option is D $3$

$f\left(x\right)={\sin }^{3}x+{\sin }^3(120+x)+{\sin }^3(240+x)$

$=\frac{3\sin x-\sin 3x+3\sin (120+x)-\sin (2\pi +3x)+3\sin (240+x)-\sin (4\pi +x)}{4}$

$[\because \sin 3x=3\sin x-4{\sin }^{3}x]$

$=\frac{3[\sin x+\sin (120+x)+\sin (240+x\left)\right]-3\sin 3x}{4}$

$=\frac{3}{4}\left[\right[\sin x+2\sin (180+x)cos{60}^o]-\sin 3x]$

$=\frac{3}{4}\left[\right[\sin x-\sin x]-\sin 3x]$

$=\frac{3}{4}[-\sin 3x]$

$f\left(x\right)=-\frac{3}{4}\sin 3x$

$\therefore$ range of $f\left(x\right)ϵ[-\frac{3}{4},\frac{3}{4}]$

So, $P=3$