Question

6sin2 x -3sinx*cosx-cos2 x-sin2 x-cos2 x=0

Answer 1

Вот пожалуйста ответ)))

Answer 2

Alright, let's tackle this equation step by step. First, let's simplify it:

$6\sin^2 x - 3\sin x \cos x - \cos^2 x - \sin^2 x - \cos^2 x = 0$

Combine like terms:

$6\sin^2 x - 4\cos^2 x - 3\sin x \cos x - \sin^2 x = 0$

Now, we can try to factor it or simplify further. Notice that we have both $\sin^2 x$ and $\cos^2 x$, which can be simplified using the trigonometric identity $\sin^2 x + \cos^2 x = 1$.

$6(1 - \cos^2 x) - 4\cos^2 x - 3\sin x \cos x - \sin^2 x = 0$

Distribute the 6:

$6 - 6\cos^2 x - 4\cos^2 x - 3\sin x \cos x - \sin^2 x = 0$

Combine like terms:

$-10\cos^2 x - 3\sin x \cos x - \sin^2 x + 6 = 0$

Now, the equation looks a bit messy. If there's a specific solution you're looking for or a range of values for $x$, let me know!