Question

Sin2x=sin(x+п/2).6соs^2x-7cosx-5=0

Answer 1

Ответ:

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Объяснение:

Answer 2

It seems like you've provided two separate equations. Let's solve them one by one.

Equation 1: $\sin(2x) = \sin\left(x + \frac{\pi}{2}\right)$

We can use trigonometric identities to simplify this equation:

$\sin(2x) = \cos(x)$

Now, we need to solve for $x$ within a certain range (usually $0 \leq x < 2\pi$ or $0 \leq x < 360^\circ$).

Using the fact that $\sin(2x) = 2\sin(x)\cos(x)$ and substituting, we get:

$2\sin(x)\cos(x) = \cos(x)$

Dividing both sides by $\cos(x)$, we get:

$2\sin(x) = 1$

Now, solve for $x$:

$\sin(x) = \frac{1}{2}$

This occurs at two angles: $x = \frac{\pi}{6}$ and $x = \frac{5\pi}{6}$, or in degrees, $x = 30^\circ$ and $x = 150^\circ$.

Equation 2: $6\cos^2(x) - 7\cos(x) - 5 = 0$

This is a quadratic equation in terms of $\cos(x)$. Let's solve it using the quadratic formula:

The quadratic formula states that for an equation of the form $ax^2 + bx + c = 0$, the solutions for $x$ are given by:

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

In your equation, $a = 6$, $b = -7$, and $c = -5$. Plug these values into the formula:

$\cos(x) = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 6 \cdot (-5)}}{2 \cdot 6}$

$\cos(x) = \frac{7 \pm \sqrt{49 + 120}}{12}$

$\cos(x) = \frac{7 \pm \sqrt{169}}{12}$

$\cos(x) = \frac{7 \pm 13}{12}$

So, you have two possible solutions:

$\cos(x) = \frac{20}{12} \quad \text{or} \quad \cos(x) = -\frac{6}{12}$

Simplifying these fractions:

$\cos(x) = \frac{5}{3} \quad \text{or} \quad \cos(x) = -\frac{1}{2}$

However, the cosine function only takes values between -1 and 1, so the second solution $\cos(x) = -\frac{1}{2}$ is valid.

This occurs at an angle $x$ where $\cos(x) = -\frac{1}{2}$. Such angles are $x = \frac{2\pi}{3}$ and $x = \frac{4\pi}{3}$, or in degrees, $x = 120^\circ$ and $x = 240^\circ$.