Question

Solve the following inequality:
$\sqrt{5-2\sin x}⩾6\sin x-1$

Answer 1

Let $6sinx-1\ge 0$

$sinx\ge 1/6$

$xϵ[2n\pi +{\sin }^{-1}\left(1/6\right),(2n+1)\pi -{\sin }^{-1}\left(1/6\right)]$

For this condition,

$\to 5-2sinx\ge {(6sinx-1)}^2$

$\to 5-2sinx\ge 36{\sin }^{2}x-12sinx+1$

$\to 36{\sin }^{2}x-10sinx-4\le 0$

$\to 36{\sin }^{2}x-10sinx-4\le 0$

Let $sinx=t$

$\to 36t^2-10t-4\le 0$

$\to 36t^2-18t+8t-4\le 0$

$\to 18t(2t-1)+4(2t-1)\le 0$

$\to (18t+4)(2t-1)\le 0$

$tϵ[\frac{-4}{18},\frac{1}{2}]$

$tϵ[\frac{-2}{9},\frac{1}{2}]$

for $t=sinx$

$sinxϵ[-2/9,\frac{1}{2}]$

So,

$xϵ[2n\pi -{\sin }^{-1}(+2/9),2n\pi +\pi /6]U$

$[(2n+1)\pi -\pi /6,(2n+1)\pi +{\sin }^{-1}\left(2/9\right)]$

Combining this and condition for

$sinx\ge 1/6$, we get

$xϵ[2n\pi +{\sin }^{-1}\left(1/6\right),2n\pi +\pi /6]U[(2n+1)\pi -\frac{\pi }{6},(2n+1)\pi -{\sin }^{-1}\left(1/6\right)]$

Now, when $sinx\le 1/6$

$\to {\left(\sqrt{5-2sinx}\right)}^2\ge 0$

$\to 5-2sinx\ge 0$

$\to sinx\le 5/2$

$xϵR$

and as $sinx\le 1/6$

$xϵ[2n\pi +{\sin }^{-1}\left(1/6\right),(2n-1)\pi -{\sin }^{-1}\left(1/6\right)]$