Question

Cos^2+cosx=sin^2x.найдите наибольший отрицательный его корень.

Answer 1

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Answer 2

To find the largest negative root of the equation $\cos^2(x) + \cos(x) = \sin^2(x)$, we need to solve the equation and then identify the negative roots.

Let's rearrange the equation: $\cos^2(x) + \cos(x) - \sin^2(x) = 0$

Now, we can use the trigonometric identity $\sin^2(x) + \cos^2(x) = 1$ to rewrite the equation as: $\cos^2(x) + \cos(x) - (1 - \cos^2(x)) = 0$

Simplifying further: $\cos^2(x) + \cos(x) - 1 + \cos^2(x) = 0$ $2\cos^2(x) + \cos(x) - 1 = 0$

Now, this is a quadratic equation in terms of $\cos(x)$. Let's solve it using the quadratic formula: $\cos(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our equation, $a = 2$, $b = 1$, and $c = -1$.

$\cos(x) = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}$ $\cos(x) = \frac{-1 \pm \sqrt{9}}{4}$ $\cos(x) = \frac{-1 \pm 3}{4}$

Now, we have two possible values for $\cos(x)$: $\cos(x) = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad \cos(x) = \frac{-4}{4} = -1$

However, since the original equation is $\cos^2(x) + \cos(x) = \sin^2(x)$, the second root $\cos(x) = -1$ is extraneous because it would make the left side of the equation equal to zero. Thus, the valid solution is $\cos(x) = \frac{1}{2}$.

Now, we need to find the corresponding value of $x$ for $\cos(x) = \frac{1}{2}$. We know that $\cos(x) = \frac{1}{2}$ when $x$ is $\frac{\pi}{3}$ or $\frac{5\pi}{3}$.

However, since we are looking for the largest negative root, we'll consider the value $x = \frac{5\pi}{3}$. This corresponds to a value of $\cos(x) = \frac{1}{2}$ and is negative, as $\cos\left(\frac{5\pi}{3}\right) = \frac{1}{2}$ and $\frac{5\pi}{3}$ lies in the second quadrant.

So, the largest negative root of the equation is $x = \frac{5\pi}{3}$.