2tgx*cosx-2cosx-tgx+1=0

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To solve the equation 2tg(x)cos(x) - 2cos(x) - tg(x) + 1 = 0, we can simplify by using the trigonometric identities:

tan(x) = sin(x)/cos(x) and cos(x) = 1/sqrt(1 + tan^2(x))

Substitute these identities into the equation:

2(sin(x)/cos(x))(1/sqrt(1 + tan^2(x))) - 2(1/sqrt(1 + tan^2(x))) - (sin(x)/cos(x)) + 1 = 0

Simplify further:

(2sin(x)/sqrt(cos^2(x) + sin^2(x))) - (2/sqrt(cos^2(x) + sin^2(x))) - (sin(x)/cos(x)) + 1 = 0

Now, let's substitute sin(x) = y and cos(x) = x:

(2y/sqrt(1)) - (2/sqrt(1)) - (y/x) + 1 = 0

2y - 2 - y + x = 0

Combine like terms:

y + x - 2 = 0

Since y = sin(x) and x = cos(x), we have sin(x) + cos(x) - 2 = 0

So the final solution to the equation 2tg(x)cos(x) - 2cos(x) - tg(x) + 1 = 0 is sin(x) + cos(x) - 2 = 0.