Ctg²(x-π/2)-ctg(x-3π/2)-2=0

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To solve the equation ctg²(x-π/2) - ctg(x-3π/2) - 2 = 0, we can use the fact that ctg(x) = 1/tan(x).

Let's start by substituting ctg(x) with 1/tan(x) in the equation:

(1/tan(x-π/2))² - 1/tan(x-3π/2) - 2 = 0

Now, let's simplify this expression:

(1/tan(x-π/2))² - 1/tan(x-3π/2) - 2 = 0

(1/(tan(x-π/2))² = 1/(tan(x-π/2))^2 = 1/(cot(x))^2 = cot(x)^2

cot(x)^2 - 1/tan(x-3π/2) - 2 = 0

Now, we can substitute tan(x-3π/2) with -cot(x) in the equation:

cot(x)^2 - 1/(-cot(x)) - 2 = 0

cot(x)^2 + cot(x) - 2 = 0

Now, we have a quadratic equation in terms of cot(x). Let's solve it by factoring or using the quadratic formula:

Using the quadratic formula:

cot(x) = [-1 ± √(1² - 4*(-2))]/2
cot(x) = [-1 ± √(1 + 8)]/2
cot(x) = (-1 ± √9)/2
cot(x) = (-1 ± 3)/2

This gives us two possible values for cot(x):

Therefore, the solutions for the equation ctg²(x-π/2)-ctg(x-3π/2)-2=0 are x = arccot(1) and x = arccot(-2), where arccot is the inverse cotangent function.