Question

Evaluate: $\frac{2sin2x-cosx}{6-{cos}^{2}x-4sinx}$

Answer 1

We have,

$\frac{2\sin 2x-\cos x}{6-{\cos }^{2}x-4\sin x}$

Then,

$\frac{2\left(2\sin x\cos x\right)-\cos x}{6-(1-{\sin }^{2}x)-4\sin x}$

$\Rightarrow \frac{4\sin x\cos x-\cos x}{6-(1-{\sin }^{2}x)-4\sin x}$

$\Rightarrow \frac{4\sin x\cos x-\cos x}{6-1+{\sin }^{2}x-4{\sin }^{2}x}$

$\Rightarrow \frac{\cos x(4\sin x-1)}{5-3{\sin }^{2}x}$

Hence, this is the answer.