If 4cos^2θ-3sin^2θ + 2 =0, then the value of tanθ is (where 0 ≤ θ ≤ 9…

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  = \(\sqrt {6 }\)