If $$3 - 2 \sin^2 \theta - 3 \cos \theta = 0, 0^\circ < \theta < 90^\circ$$, then what is the value of $$(2 \cosec \theta + \tan \theta)$$ ?

Given that,
$$3 - 2 \sin^2 \theta - 3 \cos \theta = 0$$
$$1+2 - 2 \sin^2 \theta - 3 \cos \theta = 0$$
$$1+2(1-sin^2\theta)-3 \cos \theta=0$$

$$1+2(\cos^2\theta)-3 \cos \theta=0$$

$$2\cos^2\theta-3\cos\theta+1=0$$
let $$\cos\theta=x$$
So, $$2x^2-3x+1=0$$
$$2x^2-2x-x+1=0$$
$$2x(x-1)-1(x-1)=0$$
$$(x-1)(2x-1)=0$$

So, $$X=1$$ and $$X=1/2$$

Hence, $$\cos\theta=\cos0$$ or $$\cos\theta=\cos60$$
Hence, $$\theta=0$$ or $$\theta=60^o$$

But, $$0^\circ < \theta < 90^\circ$$ so, here $$\theta=0$$ will be neglected.
Now, substituting the value $$\theta=60^o$$
$$(2 \cosec \theta + \tan \theta)$$
$$(2 \cosec 60 + \tan 60)=\dfrac{4}{\sqrt3}+\sqrt3=\dfrac{7}{\sqrt3}$$
$$Now multiply and divide by \sqrt3 $$
$$=\dfrac{7\times{\sqrt3}}{\sqrt3\times\sqrt3}$$
$$=\dfrac{7\sqrt3}{3}$$