The minimum value of α for which the equation 4 sinx+ 1 1- sinx=α has at least one solution in (0, π 2) is.

Let

$$f(x)=\frac{4}{\mathrm{sin}x}+\frac{1}{1-\mathrm{sin}x}$$f(x)=sinx4​+1−sinx1​

$$\Rightarrow f(x)=\frac{4}{\mathrm{sin}x}+\frac{1+\mathrm{sin}x}{1-{\mathrm{sin}}^{2}x}$$⇒f(x)=sinx4​+1−sin2x1+sinx​

$$\Rightarrow f(x)=4\mathrm{cosec}x+{\mathrm{sec}}^{2}x+\mathrm{tan}x\mathrm{sec}x$$⇒f(x)=4cosecx+sec2x+tanxsecx

$$\Rightarrow {f}^{'}(x)=\frac{-4\mathrm{cos}x}{{\mathrm{sin}}^{2}x}+\frac{2\mathrm{sin}x+1}{{\mathrm{cos}}^{3}x}+\frac{{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{3}x}$$⇒f′(x)=sin2x−4cosx​+cos3x2sinx+1​+cos3xsin2x​

$$\Rightarrow {f}^{'}(x)=\frac{-4\mathrm{cos}x}{{\mathrm{sin}}^{2}x}+\frac{{(\mathrm{sin}x+1)}^{2}}{{\mathrm{cos}}^{3}x}$$⇒f′(x)=sin2x−4cosx​+cos3x(sinx+1)2​

$$\Rightarrow {f}^{''}(x)=-4[\frac{-{\mathrm{sin}}^{3}x-2\mathrm{sin}x{\mathrm{cos}}^{2}x}{{\mathrm{sin}}^{4}x}]+\frac{2(\mathrm{sin}x+1){\mathrm{cos}}^{4}x+3{\mathrm{cos}}^{2}x\mathrm{sin}x{(\mathrm{sin}x+1)}^{2}}{{\mathrm{cos}}^{6}x}$$⇒f′′(x)=−4[sin4x−sin3x−2sinxcos2x​]+cos6x2(sinx+1)cos4x+3cos2xsinx(sinx+1)2​

$$\Rightarrow {f}^{''}(x)=-4[\frac{-{\mathrm{sin}}^{3}x-2\mathrm{sin}x(1-{\mathrm{sin}}^{2}x)}{{\mathrm{sin}}^{4}x}]+\frac{2(\mathrm{sin}x+1){(1-{\mathrm{sin}}^{2}x)}^{2}+3(1-{\mathrm{sin}}^{2}x)\mathrm{sin}x{(\mathrm{sin}x+1)}^{2}}{{(1-{\mathrm{sin}}^{2}x)}^{3}}$$⇒f′′(x)=−4[sin4x−sin3x−2sinx(1−sin2x)​]+(1−sin2x)32(sinx+1)(1−sin2x)2+3(1−sin2x)sinx(sinx+1)2​

For critical points,

$${f}^{'}(x)=0$$f′(x)=0

$$\Rightarrow \frac{{(\mathrm{sin}x+1)}^{2}}{{\mathrm{cos}}^{3}x}=\frac{4\mathrm{cos}x}{{\mathrm{sin}}^{2}x}$$⇒cos3x(sinx+1)2​=sin2x4cosx​

$$\Rightarrow {\mathrm{sin}}^{2}x{(\mathrm{sin}x+1)}^{2}=4{\mathrm{cos}}^{4}x$$⇒sin2x(sinx+1)2=4cos4x

$$\Rightarrow \mathrm{sin}x(\mathrm{sin}x+1)=2{\mathrm{cos}}^{2}x$$⇒sinx(sinx+1)=2cos2x

$$\Rightarrow {\mathrm{sin}}^{2}x+\mathrm{sin}x=2-2{\mathrm{sin}}^{2}x$$⇒sin2x+sinx=2−2sin2x

$$\Rightarrow 3{\mathrm{sin}}^{2}x+\mathrm{sin}x-2=0$$⇒3sin2x+sinx−2=0

$$\Rightarrow 3{\mathrm{sin}}^{2}x+3\mathrm{sin}x-2\mathrm{sin}x-2=0$$⇒3sin2x+3sinx−2sinx−2=0

$$\Rightarrow (3\mathrm{sin}x-2)(\mathrm{sin}x+1)=0$$⇒(3sinx−2)(sinx+1)=0

$$\Rightarrow \mathrm{sin}x=\frac{2}{3}$$⇒sinx=32​

Since, $\mathrm{sin}x>0\forall x\in (0,\frac{\pi }{2})$sinx>0∀x∈(0,2π​)

Now, at $\mathrm{sin}x=\frac{2}{3}$sinx=32​, we get

$$\Rightarrow {f}^{''}(x)=-4[\frac{-\frac{8}{27}-\frac{4}{3}(1-\frac{4}{9})}{\frac{16}{81}}]+\frac{2\times \frac{5}{3}{(1-\frac{4}{9})}^{2}+3(1-\frac{4}{9})\times \frac{2}{3}\times \frac{25}{9}}{{(1-\frac{4}{9})}^{3}}$$⇒f′′(x)=−4[8116​−278​−34​(1−94​)​]+(1−94​)32×35​(1−94​)2+3(1−94​)×32​×925​​

$$\Rightarrow {f}^{''}(x)=[4[\frac{28}{27}\times \frac{81}{16}]+\frac{2\times \frac{5}{3}{(\frac{5}{9})}^{2}+3(\frac{5}{9})\times \frac{2}{3}\times \frac{25}{9}}{{(\frac{5}{9})}^{3}}]>0$$⇒f′′(x)=[4[2728​×1681​]+(95​)32×35​(95​)2+3(95​)×32​×925​​]>0

Hence, it is the point of minima.

$$\therefore f{(x)}_{\mathrm{min}}=\frac{4}{\frac{2}{3}}+\frac{1}{1-\frac{2}{3}}=9$$∴f(x)min​=32​4​+1−32​1​=9

$$f{(x)}_{\mathrm{max}}\rightarrow \infty$$f(x)max​→∞

$f(x)$f(x) is continuous function

$$\therefore {\alpha }_{\mathrm{min}}=9$$∴αmin​=9