If 4 - 2sin²θ - 5cosθ = 0, 0°

If 4 - 2sin²θ - 5cosθ = 0, 0° Correct Answer $$\frac{{3\sqrt 3 }}{2}$$

$$\eqalign{ & 4 - 2{\sin ^2}\theta - 5\cos \theta = 0 \cr & {\text{Let }}\theta = {60^ \circ } \cr & 4 - 2{\sin ^2}{60^ \circ } - 5\cos {60^ \circ } = 0 \cr & 4 - 2 \times \frac{3}{4} - 5 \times \frac{1}{2} = 0 \cr & 4 - 4 = 0 \cr & \sin \theta + \tan \theta = \sin {60^ \circ } + \tan {60^ \circ } \cr & = \frac{{\sqrt 3 }}{2} + \sqrt 3 \cr & = \frac{{3\sqrt 3 }}{2} \cr} $$