Question

Find the least positive angle measured in degrees satisfying the equation :
${\sin }^{3}x+{\sin }^{3}2x+{\sin }^{3}3x=(\sin x+\sin 2x+\sin 3x{)}^3$.

• A. ${72}^{\circ }$
• B. ${36}^{\circ }$
• C. ${54}^{\circ }$
• D. None of these

Answer 1

The correct option is A ${72}^{\circ }$

Given, ${\sin }^{3}x+{\sin }^{3}2x+{\sin }^{3}3x={(\sin x+\sin 2x+\sin 3x)}^3\Rightarrow -3(\sin x+\sin 2x)(\sin x+\sin 3x)(\sin 2x+\sin 3x)=0$
$\Rightarrow (\sin x+\sin 2x)=0$ or $(\sin x+\sin 3x)=0$ or $(\sin 2x+\sin 3x)=0$

Then $(\sin x+\sin 2x)=0\Rightarrow \sin x+2\sin x\cos x=0\Rightarrow \sin x(1+2\cos x)=0$
$\Rightarrow x=n\pi ,\frac{2\pi }{3}+2n\pi ,\frac{4\pi }{3}+2n\pi$
$(\sin x+\sin 3x)=0\Rightarrow 4\sin x-4{\sin }^{3}x=0\Rightarrow \sin x(\sin x-1)(\sin x+1)=0$
$\Rightarrow x=\frac{\pi }{2}+2n\pi ,2n\pi -\frac{\pi }{2},n\pi$

$(\sin 2x+\sin 3x)=0$
Put $y=\tan \frac{x}{2}$

$\frac{6y}{{(y^2+1)}^3}-\frac{20y^3}{{(y^2+1)}^3}+\frac{6y^5}{{(y^2+1)}^3}+\frac{4y}{{(y^2+1)}^2}-\frac{4y^2}{{(y^2+1)}^2}=0$
$\Rightarrow \frac{2(y^5-10y^3+5y)}{{(y^2+1)}^3}=0\Rightarrow y(y^4-10y^2+5)=0$
$\Rightarrow y=0$ or $(y^4-10y^2+5)=0$

Solving this
$\Rightarrow x=2n\pi ,\frac{4\pi }{5}+2n\pi ,2n\pi -\frac{4\pi }{5},\frac{2\pi }{5}+2n\pi$
Hence
$x=2n\pi ,2n\pi +\pi ,\frac{2}{3}(3n\pi -\pi ),\frac{2}{3}(3n\pi +\pi ),\frac{1}{2}(4n\pi -\pi )$

Therefore least value is $x={72}^0$