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.
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.