(1+cos2x)sinx=cos^2xsin3x=-2cosxsin3x-2sinxcos2x=0

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вопрос

To solve this trigonometric equation, we will start by expressing everything in terms of sine and cosine functions.

Given equation: (1 + cos(2x))sin(x) = cos^2(x)sin(3x) = -2cos(x)sin(3x) - 2sin(x)cos(2x) = 0

First, let's simplify the terms:

(1 + cos(2x))sin(x) = sin(x) + cos(2x)sin(x) = sin(x) + 2sin(x)cos(x) = sin(x) + sin(2x)

cos^2(x)sin(3x) = cos(x)cos(x)sin(3x) = cos(x)sin(3x)cos(x) = sin(3x)cos(2x)

-2cos(x)sin(3x) - 2sin(x)cos(2x) = -2cos(x)sin(3x) - 2sin(x)cos(2x) = -2sin(3x) - 2sin(2x) = -2sin(3x) - 2(2sin(x)cos(x)) = -2sin(3x) - 4sin(x)cos(x) = sin(3x) - 4sin(x)cos(x) = 0

Now, we have the equation in terms of sine and cosine functions:

sin(x) + sin(2x) = sin(3x) - 4sin(x)cos(x)

Rearranging terms:

sin(x) - 4sin(x)cos(x) = sin(3x) - sin(2x)

Using trigonometric identities, we can simplify this equation further, but it may not be possible to solve it completely without numerical methods or graphing techniques.