Question

2 sin2 x − sin x cos x − cos2 x = 0;

Answer 1

Ответ:

Пошаговое объяснение:

делим все на cos^2x

получим:

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Answer 2

To solve the equation $2\sin^2 x - \sin x \cos x - \cos^2 x = 0$, you can use some trigonometric identities to simplify it. Here's the step-by-step solution:

First, notice that $\sin^2 x + \cos^2 x = 1$ is the Pythagorean identity. We can use this identity to simplify the equation:

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

Now, replace $\cos^2 x$ with $1 - \sin^2 x$:

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

Now, distribute the negative sign on the right side:

$$2\sin^2 x - \sin x \cos x - 1 + \sin^2 x = 0$$

Combine like terms:

$$3\sin^2 x - \sin x \cos x - 1 = 0$$

Now, let's make a substitution to simplify the equation further. Let $u = \sin x$, so $u^2 = \sin^2 x$. We can rewrite the equation as:

$$3u^2 - u\cos x - 1 = 0$$

This is now a quadratic equation in terms of $u$. You can solve it using the quadratic formula:

$$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

In this case, $a = 3$, $b = -\cos x$, and $c = -1$. Plugging these values into the formula:

$$u = \frac{-(-\cos x) \pm \sqrt{(-\cos x)^2 - 4(3)(-1)}}{2(3)}$$

Now, simplify further:

$$u = \frac{\cos x \pm \sqrt{\cos^2 x + 12}}{6}$$

So, you have two possible solutions for $u$:

1. $u_1 = \frac{\cos x + \sqrt{\cos^2 x + 12}}{6}$
2. $u_2 = \frac{\cos x - \sqrt{\cos^2 x + 12}}{6}$

Now, remember that $u = \sin x$, so you can use these solutions for $u$ to find the corresponding solutions for $x$:

1. $x_1 = \arcsin\left(\frac{\cos x + \sqrt{\cos^2 x + 12}}{6}\right)$
2. $x_2 = \arcsin\left(\frac{\cos x - \sqrt{\cos^2 x + 12}}{6}\right)$

These are the solutions to the original equation $2\sin^2 x - \sin x \cos x - \cos^2 x = 0$. Depending on the values of $\cos x$ and whether they fall within the domain of the arcsin function, you can determine the specific solutions for $x$.