Question

Sin x < 1/2cos x > √2/2tg x < 1​

Answer 1

Ответ:

1.

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

рисунок1

$x \in( - \frac{7\pi}{6} + 2\pi \: n; \frac{\pi}{6} + 2\pi \: n)$

2.

$\cos(x) > \frac{ \sqrt{2} }{2}$

рисунок 2

$x \in( - \frac{\pi}{4} + 2\pi \: n; \frac{\pi}{4} + 2\pi \: n)$

3.

$tg(x) < 1$

рисунок 3

$x \in( - \frac{\pi}{2} + \pi \: n; \frac{\pi}{4} + \pi \: n)$

везде n принадлежит Z.

Answer 2

To solve the inequalities:

1. $\sin(x) < \frac{1}{2}$
2. $\cos(x) > \frac{\sqrt{2}}{2}$
3. $\tan(x) < 1$

Let's solve them one by one:

1. $\sin(x) < \frac{1}{2}$

To solve this inequality, we can use the unit circle or the graph of the sine function. On the unit circle, the sine function represents the y-coordinate of the point corresponding to the angle $x$.

The sine function is positive in the first and second quadrants and negative in the third and fourth quadrants. To find where $\sin(x) < \frac{1}{2}$, we need to look for angles where the y-coordinate is less than $\frac{1}{2}$.

So, in the first quadrant, $0 < x < \frac{\pi}{6}$ satisfies $\sin(x) < \frac{1}{2}$.

In the second quadrant, $\frac{\pi}{6} < x < \frac{\pi}{2}$ also satisfies $\sin(x) < \frac{1}{2}$.

In the third quadrant, $\pi < x < \frac{7\pi}{6}$ also satisfies $\sin(x) < \frac{1}{2}$.

In the fourth quadrant, $\frac{7\pi}{6} < x < 2\pi$ satisfies $\sin(x) < \frac{1}{2}$.

So, the solutions to $\sin(x) < \frac{1}{2}$ are: $0 < x < \frac{\pi}{6} \text{ or } \frac{\pi}{6} < x < \frac{\pi}{2} \text{ or } \pi < x < \frac{7\pi}{6} \text{ or } \frac{7\pi}{6} < x < 2\pi$

2. $\cos(x) > \frac{\sqrt{2}}{2}$

Similar to the sine inequality, we can use the unit circle or the graph of the cosine function to find where $\cos(x) > \frac{\sqrt{2}}{2}$.

The cosine function is positive in the first and fourth quadrants and negative in the second and third quadrants. To find where $\cos(x) > \frac{\sqrt{2}}{2}$, we need to look for angles where the x-coordinate is greater than $\frac{\sqrt{2}}{2}$.

In the first quadrant, $0 < x < \frac{\pi}{4}$ satisfies $\cos(x) > \frac{\sqrt{2}}{2}$.

In the fourth quadrant, $\frac{7\pi}{4} < x < 2\pi$ also satisfies $\cos(x) > \frac{\sqrt{2}}{2}$.

So, the solutions to $\cos(x) > \frac{\sqrt{2}}{2}$