Question

$\mathrm{The}\mathrm{general}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{equation}\sqrt{5-2\sin x}=6\sin x-1\mathrm{is}\mathrm{given}\mathrm{by}$.

• A. $\mathrm{n\pi }+{(-1)}^n\frac{\pi }{6},n\in Z$
• B. $\mathrm{n\pi }\pm \frac{\pi }{6},n\in Z$
• C. $\mathrm{n\pi }+{(-1)}^n\frac{\pi }{4},n\in Z$

Answer 1

The correct option is A $\mathrm{n\pi }+{(-1)}^n\frac{\pi }{6},n\in Z$

We have the given equation:
$\sqrt{5-2\mathrm{sinx}}=6\mathrm{sinx}-1$.
Now, the equation is meaningful if $\mathrm{sinx}\le \frac{5}{2}$ which is always true.
Also, $6\sin x-1>0$ as L.H.S is always positive.
Now,
$6\sin x-1>0\Rightarrow \sin x>\frac{1}{6}.$
On squaring the equation, we get:
$5-2\sin x={\left(6\sin x-1\right)}^2\Rightarrow 5-2\sin x=\left(36{\sin }^{2}x+1-12\sin x\right)\Rightarrow 36{\sin }^{2}x-10\sin x-4=0\Rightarrow 18{\sin }^{2}x-5\sin x-2=0\Rightarrow (9\sin x+2)(2\sin x-1)=0\Rightarrow \sin x=\frac{-2}{9}\mathrm{or}\sin x=\frac{1}{2}\mathrm{But},\sin x=-\frac{2}{9}\mathrm{is}\mathrm{not}\mathrm{possible}\mathrm{as}\sin x>\frac{1}{6}.\mathrm{Thus},\mathrm{the}\mathrm{only}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{equation}\mathrm{is}:\sin x=\frac{1}{2}=\sin \frac{\pi }{6}\Rightarrow x=n\pi +{(-1)}^n\frac{\pi }{6},n\in Z.$