What is the sum of the solutions of tan(3x - π/6) = cot(x²) in the interval [-3π/2, π/2]?

The sum is -3. You don't have to actually find the roots to get this.

Using the cofunction identity cot(Θ) = tan(π/2 - Θ) gives

tan(3x - π/6) = tan(π/2 - x²)

Using the π-periodicity of the tangent function, this gives the general equation

tan(3x - π/6) = tan(π/2 - x² + kπ) ==> 3x - π/6 = π/2 - x² + kπ ==>

x² + 3x - 2π/3 - kπ = 0

where k is any integer. For any fixed value of k, this quadratic has two roots x1 and x2. Turns out taking k = 0 gives these roots between -3π/2 and π/2.

The quadratic can be written as

x² + 3x - 2π/3 = (x - x1)(x - x2) = 0 ==> -(x1 + x2) = 3 and x1 x2 = -2π/3

So x1 + x2 = -3.

***P.S. ****

If you take any other value of k, you'll still get the sum of the roots to be -3. The product of the roots will change though.

cot(2pi/3 - 3x) = cot(x^2)

x^2 = 2pi/3 - 3x

x^2 + 3x - 2pi/3 = 0

The two solutions from this equation sum to -3.

There maybe other solutions from the following equations.

x^2 + 3x + pi/3 = 0 both solutions are in the given interval and sum to -3

x^2 + 3x - 5pi/3 = 0 --> x = [-3 + sqrt(9 + 20pi/3)]/2

The total sum is -6 + 1/2 [-3 + sqrt(9+20pi/3)].

I hope i am not missing any other solutions in the given interval.

Isnt the answer -3?..Lake R is right.