Question

Sinxcosx>1/4 ????????

Answer 1

Ответ:

(см. объяснение)

Объяснение:

$sinxcosx>1/4\\sin2x>1/2\\x\in\left(\dfrac{\pi}{12}+n\pi,\;\dfrac{5\pi}{12}+n\pi\right),\;n\in Z$

Неравенство решено!

Answer 2

To determine whether the inequality sin(x)cos(x) > 1/4 holds true for all values of x, we can analyze it algebraically and graphically.

First, let's consider the inequality:

sin(x)cos(x) > 1/4

To analyze this inequality, we can use the properties of the sine and cosine functions. The maximum value of the product sin(x)cos(x) occurs when sin(x) = cos(x). In this case, sin(x)cos(x) = (sin(x))^2, and it reaches its maximum value of 1/2 when sin(x) = cos(x) = 1/sqrt(2).

So, the inequality can be rewritten as:

(sin(x))^2 > 1/4

Now, we can take the square root of both sides:

|sin(x)| > 1/2

This means that sin(x) must be greater than 1/2 or less than -1/2 for the inequality sin(x)cos(x) > 1/4 to hold.

Graphically, the sine function oscillates between -1 and 1, and the values that are greater than 1/2 or less than -1/2 are within certain intervals. So, the inequality sin(x)cos(x) > 1/4 will hold for values of x within these intervals.

In summary, sin(x)cos(x) > 1/4 is true for values of x where sin(x) is greater than 1/2 or less than -1/2. You can find these intervals by examining the graph of the sine function or by solving the inequalities involving sin(x).