Question

Sin x --------- = 0 cosx-1 решить уравнение

Answer 1

$\dfrac{\sin x}{\cos x-1}=0\\ \\ \dfrac{2\sin \frac{x}{2}\cos \frac{x}{2}}{-2\sin^2\frac{x}{2}}=0\\ \\ -\dfrac{\cos \frac{x}{2}}{\sin\frac{x}{2}}=0$

Дробь равен нулю, если числитель обращается к нулю

$\cos \frac{x}{2}=0\\ \\ \frac{x}{2}=\frac{\pi}{2}+\pi n,n \in \mathbb{Z}\\ \\ x=\pi+2\pi n,n \in \mathbb{Z}$

Ответ: $\pi +2\pi n,n \in \mathbb{Z}$

Answer 2

Дробь равна нулю, когда числитель равен нулю, а знаменатель от нуля отличен.  sinx=0 cosx≠1;  х=πn; n∈Z, из этого множества надо исключить х=2πn; n∈Z, т.е. оставить ответ нечетных n=(2к+1), окончательно х=π(2к+1); к∈Z,  х=π(2к+1); к∈Z.

Answer 3

To solve the equation (sin x) / (cos x - 1) = 0, we first need to find the values of x that satisfy this equation.

When dealing with fractions, remember that a fraction is equal to zero only if its numerator is equal to zero, and its denominator is not zero. In this case, the numerator is sin x, and the denominator is cos x - 1.

So, we set the numerator equal to zero:

sin x = 0

The sine function is equal to zero at specific angles, namely at multiples of π (pi):

x = nπ, where n is an integer.

Now, we need to check whether the denominator (cos x - 1) is not equal to zero:

cos x - 1 ≠ 0

To find the values of x where the denominator is not zero, we solve for x:

cos x ≠ 1

The cosine function is not equal to 1 for any angle. Therefore, there are no restrictions on x due to the denominator.

In conclusion, the solution to the equation (sin x) / (cos x - 1) = 0 is:

x = nπ, where n is an integer.