If $4cos^2θ-3sin^2θ + 2 =0$, then the value of tanθ is (where 0 ≤ θ ≤ 90°)

$\sqrt{6}$ 1 $\sqrt{2}$ $\frac{1}{\sqrt{3}}$

$\sqrt{6}$

4 cos²θ - 3 sin²θ + 2 = 0 { sin²θ  + cos²θ  = 1 } 4 cos²θ - 3 ( 1 - cos²θ ) + 2 = 0 7 cos²θ = 1 cos²θ = \(\frac{1 }{7}\) Now, sin²θ = 1 - \(\frac{1 }{7}\) = \(\frac{6}{7}\) Now, tan²θ = \(\frac{sin²θ }{cos²θ }\) = \(\frac{6 }{1}\) tan θ  = \(\sqrt {6 }\)