Question

Cosx+cos(п/2-x)+cos(п+x)=0

Answer 1

Ответ:

$\cos(x) + \cos( \frac{\pi}{2} - x) \cos(\pi + x) = 0 \\ \cos(x) + \sin(x) - \cos(x) = 0 \\ \sin(x) = 0 \\ x = \pi \: n \\ n\in \: Z$

Answer 2

To solve the equation $\cos(x) + \cos\left(\frac{\pi}{2} - x\right) + \cos(\pi + x) = 0$, you can use some trigonometric identities and simplifications.

1. Recall the sum-to-product trigonometric identity: $\cos(A) + \cos(B) = 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)$.

2. We can apply this identity to the first two terms: $\cos(x) + \cos\left(\frac{\pi}{2} - x\right) = 2\cos\left(\frac{x + \frac{\pi}{2} - x}{2}\right)\cos\left(\frac{x - \left(\frac{\pi}{2} - x\right)}{2}\right) = 2\cos\left(\frac{\pi}{4}\right)\cos\left(\frac{2x - \frac{\pi}{2}}{2}\right)$.

3. Simplify the third term using $\cos(\pi + x) = -\cos(x)$: $\cos(\pi + x) = -\cos(x)$.

Now, the equation becomes:

$2\cos\left(\frac{\pi}{4}\right)\cos\left(\frac{2x - \frac{\pi}{2}}{2}\right) - \cos(x) = 0$

Now, we have a trigonometric equation involving only one trigonometric function, $\cos$. Let's continue to solve it:

First, simplify $\cos\left(\frac{\pi}{4}\right)$ since $\frac{\pi}{4}$ is a known value:

$\sqrt{2}/2\cos\left(\frac{2x - \frac{\pi}{2}}{2}\right) - \cos(x) = 0$

Next, let $u = \frac{2x - \frac{\pi}{2}}{2}$ for easier manipulation. Then, $\cos(u) = \cos\left(\frac{2x - \frac{\pi}{2}}{2}\right)$.

$\sqrt{2}/2\cos(u) - \cos(x) = 0$

Now, we have the equation:

$\sqrt{2}/2\cos(u) - \cos(x) = 0$

$\sqrt{2}\cos(u) - 2\cos(x) = 0$

$\sqrt{2}\cos(u) = 2\cos(x)$

Now, solve for $\cos(x)$:

$\cos(x) = (\sqrt{2}/2)\cos(u)$

To find the solutions for $x$, you will need to solve for $u$ and then back-substitute to find $x$.

$\cos(u) = \sqrt{2}/2$

To solve for $u$, you can take the arccosine of both sides:

$u = \arccos(\sqrt{2}/2)$