2+2cosx=3sinx×cosx+2sinx

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To solve for x in the given equation 2 + 2cosx = 3sinx * cosx + 2sinx, we need to rearrange terms and simplify.

2 + 2cosx = 3sinx * cosx + 2sinx
=> 2 + 2cosx = sinx (3cosx + 2)

Now, let's use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to substitute for sin^2(x) in terms of cosx:

sin^2(x) = 1 - cos^2(x)

Substitute this into the equation:

2 + 2cosx = sinx (3cosx + 2)
=> 2 + 2cosx = sinx 3cosx + sinx 2
=> 2 + 2cosx = 3sinxcosx + 2sinx
=> 2 + 2cosx = 2sinx + 3sinxcosx

Next, use sin^2(x) + cos^2(x) = 1:

2 + 2cosx = 2sinx + 3sinxcosx
=> 2 + 2cosx = 2sinx + 3sinxcosx
=> 2 + 2cosx = 2sinx + 3sinxcosx
=> 2 + 2cosx = 2sinx + sinx(3cosx)
=> 2 + 2cosx = 2sinx + sinx(3cosx)
=> 2 + 2cosx = sinx(2 + 3cosx)

Now we have:

2 + 2cosx = sinx(2 + 3cosx)

To continue solving for x, we can square both sides to eliminate the square root:

(2 + 2cosx)^2 = sinx^2(2 + 3cosx)^2
4 + 8cosx + 4cosx^2 = sinx^2(4 + 12cosx + 9cosx^2)
4 + 8cosx + 4cosx^2 = sinx^2(4 + 12cosx + 9cosx^2)

Further simplification may involve expanding and manipulating the resulting equation algebraically to isolate the variable x. The final solution for x will be the values that satisfy the equation.