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

Given : $$3 + \cos^2 \theta = 3(\cot^2 \theta + \sin^2 \theta), 0^\circ < \theta < 90^\circ$$

=> $$3+(1-sin^2\theta)=3(cosec^2\theta-1)+3sin^2\theta$$

=> $$4-4sin^2\theta=\frac{3}{sin^2\theta}-3$$

=> $$7sin^2\theta-4sin^4\theta=3$$

Let $$sin^2\theta=x$$

=> $$4x^2-7x+3=0$$

=> $$(x-1)(4x-3)=0$$

=> $$x=1,\frac{3}{4}$$

=> $$sin\theta=1,\frac{\sqrt3}{2}$$

$$\because \theta<90^\circ$$, => $$\theta=60^\circ$$

$$\therefore$$ $$(\cos \theta + 2 \sin \theta)$$

= $$cos(60^\circ)+2sin(60^\circ)$$

= $$\frac{1}{2}+(2\times\frac{\sqrt3}{2})$$

= $$\frac{2\sqrt3+1}{2}$$

=> Ans - (C)