Question

Sin 2x < 1/2 решите неравенство cos x/3 > √3/2 решите неравенство

Answer 1

не знаю точно, как ты хочешь

Но я знаю так

1/2=П/6=30

sin2x<P/6

sinx<p/12

k3/2=30=p/6

cosx/3>p/6

cosx>p/2

Answer 2

To solve the inequalities, we need to find the values of x that satisfy each condition. Let's start with the first inequality:

1. sin(2x) < 1/2

To solve this inequality, we can use the unit circle or trigonometric identities. The unit circle values for sin(2x) are:

sin(2x) = 1/2 for x = π/6 + 2πk or x = 5π/6 + 2πk, where k is an integer.

Now, let's move on to the second inequality:

2. cos(x/3) > √3/2

To solve this inequality, we again use the unit circle or trigonometric identities. The unit circle values for cos(x/3) are:

cos(x/3) = √3/2 for x = π/6 + 2πk or x = 11π/6 + 2πk, where k is an integer.

However, the inequality is "greater than" (√3/2), not "greater than or equal to." So, we need to find the values of x that lie between the above solutions.

For this inequality, x must satisfy the following condition:

π/6 + 2πk < x < 11π/6 + 2πk, where k is an integer.

Combining the two inequalities:

x must satisfy both conditions:

1. x = π/6 + 2πk for k being any integer.
2. π/6 + 2πk < x < 11π/6 + 2πk for k being any integer.

So the solution for the combined inequalities is:

x = π/6 + 2πk, where k is an integer.