Question

Cos3x•cosx+sin3x•sinx>1/2

Answer 1

Ответ:

решение смотри на фотографии

Answer 2

To solve the inequality cos(3x) * cos(x) + sin(3x) * sin(x) > 1/2, you can use trigonometric identities to simplify it.

First, apply the product-to-sum identities:

cos(3x) * cos(x) + sin(3x) * sin(x) = cos(3x - x) = cos(2x)

Now, the inequality becomes:

cos(2x) > 1/2

To solve this inequality, consider the unit circle and the values of cosine in the first quadrant. The cosine function is positive in the first and fourth quadrants and negative in the second and third quadrants.

1. In the first quadrant (0 <= x <= π/2), cos(2x) is positive, and cos(2x) = 1 when x = 0.
2. In the fourth quadrant (3π/2 <= x <= 2π), cos(2x) is also positive, and cos(2x) = 1 when x = 2π.

So, the solutions for cos(2x) > 1/2 in the interval [0, 2π] are:

0 <= x < π/6 or 11π/6 <= x < 2π

These are the values of x that satisfy the inequality cos(3x) * cos(x) + sin(3x) * sin(x) > 1/2.