2корень2sin(x+pi/3)+2cos^2x=корень6cosx+2 Б){-3pi;-3pi/2}

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To solve this equation, we first isolate the square root on the left side of the equation:

√(2sin(x + π/3)) = √(6cosx + 2 - 2cos^2x)

Now, let's square both sides of the equation to eliminate the square root:

2sin(x + π/3) = 6cosx + 2 - 2cos^2x

Expand the right side of the equation:

2sin(x + π/3) = 6cosx + 2 - 2cos^2x
2sin(x + π/3) = 6cosx + 2 - 2(1 - sin^2x)
2sin(x + π/3) = 6cosx + 2 - 2 + 2sin^2x
2sin(x + π/3) = 6cosx + 2sin^2x

Now, let's solve for x. We have a trigonometric equation that involves both sine and cosine functions, which can be challenging to solve algebraically. One way to approach this is to use trigonometric identities to simplify the equation and find a solution.

However, if you are looking for the possible values of x in the given range (-3π, -3π/2), you can try to plug in values within that range and see if they satisfy the equation. A numerical approach might be necessary in this case.

Please let me know if you need further assistance.