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

$\frac{2\sqrt{3} + 1}{2}$

3 + cos²θ = 3 ( cot²θ + sin²θ )

Put θ = 30º

 3 + cos²θ = 3 ( cot²θ + ( 1 - cos²θ ) )

 3 + cos²θ = 3cot²θ + 3( 1 - cos²θ )

4cos²θ = 3cot²θ

Sinθ = \(\frac{√3}{2}\)

{ We know,  Sin60º = \(\frac{√3}{2}\) }

So, θ = 60º

Now,

cosθ + 2sinθ 

= cos60º + 2sin60º 

= \(\frac{1}{2}\) + 2 × \(\frac{√3}{2}\)

= \(\frac{1}{2}\)  + √3

= \(\frac{1 +2√3 }{2}\)