Question

(1+sinx)(cos2x-2)=0 решите уравнение

Answer 1

Ответ: x= - 90°+-360°

Пошаговое решение:

(1+sinx)(cos2x-2)=0

Принимая во внимание, что произведение равно 0,если один из сомножителей равен 0,решаем 2 уравнения:

1) 1+sinx=0

2) cos2x-2=0

Из 1го

sinx=-1 => x= - 90°+-360°

Из 2го

cos2x=2 - нет решения

Answer 2

Ответ: x=3π/2+2πn,

Пошаговое объяснение:

(1+sinx)(cos(2x)-2)=0

1+sinx=0

sinx=-1

x₁=3π/2+2πn

cos(2x)-2=0

cos(2x)=2  ∉, так как -1≤cos(2x)≤1.

Answer 3

To solve the equation (1 + sin(x))(cos(2x) - 2) = 0, we need to find the values of x that make the equation true. To do that, we'll set each factor to zero and solve for x:

Step 1: Set (1 + sin(x)) = 0 1 + sin(x) = 0 sin(x) = -1

Step 2: Set (cos(2x) - 2) = 0 cos(2x) - 2 = 0 cos(2x) = 2

Now, we need to find the values of x that satisfy each of these conditions.

Step 1: The range of the sine function is -1 to 1. Since sin(x) = -1 only when x = -π/2 + 2πk, where k is an integer.

Step 2: The range of the cosine function is -1 to 1. However, there are no real values of x that satisfy cos(2x) = 2, as the maximum value of the cosine function is 1.

Therefore, there are no real solutions for the equation (1 + sin(x))(cos(2x) - 2) = 0.