To solve this equation, we can use the Pythagorean identity for sine and cosine:

[ \sin^2 x + \cos^2 x = 1 ]

We can rewrite the given equation in terms of sine and cosine:

[ 4\sin^2 x - 3\sin x \cos x - \cos^2 x = 0 ]

[ \Rightarrow 4\sin^2 x - 3\sin x \cos x - (1 - \sin^2 x) = 0 ]

[ \Rightarrow 4\sin^2 x - 3\sin x \cos x - 1 + \sin^2 x = 0 ]

[ \Rightarrow 5\sin^2 x - 3\sin x \cos x - 1 = 0 ]

Now, we can substitute ( 1 - \cos^2 x ) for ( \sin^2 x ) in the equation:

[ 5(1 - \cos^2 x) - 3\sin x \cos x - 1 = 0 ]

[ \Rightarrow 5 - 5\cos^2 x - 3\sin x \cos x - 1 = 0 ]

[ \Rightarrow -5\cos^2 x - 3\sin x \cos x + 4 = 0 ]

Now, we can use the identity ( \sin x \cos x = \frac{1}{2} \sin 2x ) to simplify the equation further:

[ -5\cos^2 x - \frac{3}{2} \sin 2x + 4 = 0 ]

This is a quadratic equation in terms of ( \cos x ). We can solve it using the quadratic formula. Let ( y = \cos x ):

[ -5y^2 - \frac{3}{2} \sin 2x + 4 = 0 ]

[ 5y^2 + \frac{3}{2} \sin 2x - 4 = 0 ]

Plugging it into the quadratic formula:

[ y = \frac{- \frac{3}{2} \sin 2x \pm \sqrt{\left( \frac{3}{2} \sin 2x \right)^2 - 4 \cdot 5 \cdot ( -4 )}}{2 \cdot 5} ]

This is a complex equation, and finding the exact solution using this method may be difficult. You can simplify further and approximate the solution numerically.