Question

Cos(p/3-2x)×sin(p/3-2x)>1

Answer 1

$cos(\frac{\pi}{3}-2x)*sin(\frac{\pi}{3}-2x)>1$

=> $\frac{1}{2}sin(\frac{2\pi}{3}-4x)>1$

Область значений функции sin лежит в отрезке [-1;1]. Следовательно, неравенство не имеет решения (это мы ещё не учли коэфициент перед функцией)))

Ответ: пустое множество решений

Answer 2

This inequality cannot be true for all values of x, since the range of the sine and cosine functions is between -1 and 1, inclusive. Therefore, there must be some restriction on the values of x that satisfy the inequality.

To find this restriction, we can use the identity:

sin(a)cos(a) = 1/2 sin(2a)

where a = p/3 - 2x. Substituting this into the original inequality, we get:

1/2 sin(2a) > 1

Multiplying both sides by 2 and taking the sine inverse of both sides, we get:

2a < sin^-1(2)

a < 1/2 sin^-1(2)

Substituting back for a, we get:

p/3 - 2x < 1/2 sin^-1(2)

Solving for x, we get:

x > (1/6)p - 1/4 sin^-1(2)

Therefore, the inequality is true for all x greater than (1/6)p - 1/4 sin^-1(2).