Question

  6sin^2x+sinxcosx-cos^2x=2

Answer 1

6sin²x+sin(x)cosx-cos²x = 2·1

6sin²x+sinxcosx-cos²x = 2·(sin²x+cos²x)

4sin²x+sin(x)cosx-3cos²x = 0 |:cos²x≠0

Т.к. если cos²x=0, то sin²x=0, но это невозможно.

$4\left( \dfrac{\sin x}{\cos x} \right) ^2+\dfrac{\sin x}{\cos x} -3=0$

4tg²+tgx-3 = 0

Решим квадратное уравнение относительно tgx.

D = 1-4·4·(-3) = 1+48 = 7²

tgx = $\dfrac{-1\pm 7}{2\cdot 4} =\left \{ -1;\dfrac34 \right \}$

$\left[ \begin{array}{ccc} x={\tt arctg}\left( \dfrac34 \right) +\pi n\\x=\dfrac{-\pi }4 +\pi k\qquad \quad \end{array} \right.n,\! k\in \mathbb{Z}$ - ответ.

Answer 2

To solve the equation $6\sin^2x + \sin x \cos x - \cos^2x = 2$, you can use trigonometric identities to simplify and then solve for $x$. Here's how you can do it:

Start by using the identity $\sin^2x + \cos^2x = 1$:

$6(1 - \cos^2x) + \sin x \cos x - \cos^2x = 2$

Now, distribute the 6 on the left side:

$6 - 6\cos^2x + \sin x \cos x - \cos^2x = 2$

Rearrange the terms on the left side:

$6 - 7\cos^2x + \sin x \cos x = 2$

Subtract 2 from both sides of the equation:

$4 - 7\cos^2x + \sin x \cos x = 0$

Now, let's make a substitution. Let $u = \cos x$. Then, we have:

$4 - 7u^2 + \sin x \cdot u = 0$

Now, we need to express $\sin x$ in terms of $u$. Use the identity $\sin^2x = 1 - \cos^2x$ to solve for $\sin x$:

$\sin^2x = 1 - u^2$ $\sin x = \pm \sqrt{1 - u^2}$

Now, substitute this expression for $\sin x$ back into the equation:

$4 - 7u^2 + (\pm \sqrt{1 - u^2})u = 0$

We have two equations to solve, one with the positive sign and one with the negative sign.

1. For the positive sign:

$4 - 7u^2 + u\sqrt{1 - u^2} = 0$

2. For the negative sign:

$4 - 7u^2 - u\sqrt{1 - u^2} = 0$

These are quadratic equations in $u$, and you can solve them using standard methods. Once you find the solutions for $u$, you can then find the corresponding values of $x$ by taking the inverse cosine:

$x = \arccos(u)$

Please note that these equations may have multiple solutions depending on the value of $u$.