1) Sin(4x-П/6)=02) sin^2x-2sinxCosx=3cos^2x

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вопрос

1) To solve sin(4x-π/6) = 0, we need to find the values of x which make the expression equal to 0.

sin(4x-π/6) = 0
4x-π/6 = kπ where k is an integer
4x = kπ + π/6
x = (kπ + π/6) / 4

So, x = (kπ + π/6) / 4 where k is an integer.

2) To solve sin^2x - 2sinx*cosx = 3cos^2x, we can rewrite the equation in terms of sine and cosine.

sin^2x - 2sinxcosx = 3cos^2x
sin^2x - 2sinxcosx = 3(1 - sin^2x) (using the identity cos^2x = 1 - sin^2x)
sin^2x - 2sinx*cosx = 3 - 3sin^2x
4sin^2x - 2sinx - 3 = 0

This quadratic equation can be solved to find the values of sinx and then obtain the corresponding values of x.