Question

Consider the equation $\sin 3\alpha =4\sin \alpha .\sin (x+\alpha )\sin (x-\alpha ),\alpha ϵ(0,\pi )$ and $xϵ(0,2\pi )$, then :

• A. equation has no solution in II quadrant
• B. equation has no solution in III quadrant
• C. equation has one solution in each quadrant
• D. none of these

Answer 1

Answer: (C)

equation has one solution in each quadrant

$\sin 3\alpha =4\sin \alpha .\sin (x+\alpha )\sin (x-\alpha )3\sin \alpha -4{\sin }^3\alpha =4\sin \alpha .\sin (x+\alpha )\sin (x-\alpha )\Rightarrow 3-4{\sin }^2\alpha =4({\sin }^{2}x-{\sin }^2\alpha )$
$\Rightarrow 3=4{\sin }^{2}x\Rightarrow {\sin }^{2}x=\frac{3}{4}$
$={\left(\frac{\sqrt{3}}{2}\right)}^2={\sin }^2(\pm \frac{\pi }{3})$
$\Rightarrow x=k\pi \pm \frac{\pi }{3}$