2sinx-cos^2*x*sinx=0

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вопрос

To solve the equation 2sinx - cos^2(x)sinx = 0, we can factor out a sinx term:

sinx(2 - cos^2(x)) = 0

Now, we have two possibilities for sinx to be equal to 0 or for (2 - cos^2(x)) to be equal to 0:

1) sinx = 0
This implies x = nπ, where n is an integer.

2) 2 - cos^2(x) = 0
cos^2(x) =
cos(x) = ±√2

The solutions for cos(x) = ±√2 are not in the range of values for the cosine function (-1 <= cos(x) <= 1). Therefore, there are no solutions for this part of the equation.

In conclusion, the solution to the equation 2sinx - cos^2(x)sinx = 0 is x = nπ, where n is an integer.