Question

Cosxsinπ/8-sinxcosπ/8<0

Answer 1

cos(x)sin(π/8)-sin(x)cos(π/8)<0

sin(π/8-x)<0

Найдём нули

sin(π/8-x)=0<=>π/8-x=πk=>x=π/8+πk

π/8+πk<x<9π/8+πk

k пренадлежит Z

Answer 2

To solve the inequality $\cos(x) \sin(\pi/8) - \sin(x) \cos(\pi/8) < 0$, we can use trigonometric identities to simplify it.

First, let's use the trigonometric identity for the sine of the sum of two angles:

$\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$.

In this case, we can rewrite the inequality as:

$\sin\left(x + \frac{\pi}{8}\right) < 0$.

Now, we need to determine the values of $x$ that make $\sin\left(x + \frac{\pi}{8}\right)$ negative. The sine function is negative in the second and third quadrants of the unit circle.

1. In the second quadrant, $\pi/2 < x + \pi/8 < \pi$. So, $\pi/8 < x < 7\pi/8$.

2. In the third quadrant, $\pi < x + \pi/8 < 3\pi/2$. So, $7\pi/8 < x < 15\pi/8$.

So, the solution to the inequality is:

$\frac{\pi}{8} < x < \frac{7\pi}{8}$ or $x > \frac{15\pi}{8}$.

These are the values of $x$ that make $\cos(x) \sin(\pi/8) - \sin(x) \cos(\pi/8) < 0$.