Solution Set for cot-1(x)2 -(5 cot-1(x)) +6 >0?

[takando12]

Homework Statement

Solution set of the inequality (cot^-1(x))² -(5 cot^-1(x)) +6 >0 is?

Homework Equations

The Attempt at a Solution

Subs cot^-1(x)=y
We get a quadratic inequality in y.
y²-5y+6>0
(y-2)(y-3)>0
Using the wavy curve method, the solution set is,
y∈(-∞,2) ∪(3,∞)
So cot^-1(x)<2 and cot^-1(x)>3
Taking cot on both sides of the inequality,
x<cot2 and x>cot3
x∈(-∞,cot2) ∪(cot3,∞)
Yet the answer is (-∞,cot3)∪(cot2,∞).
I'm guessing that in the step where I take cot on both sides, I'll have to change the inequality signs as arccot is a decreasing function. Is that where the problem lies?

 

[tommyxu3]

An idea: I think you have to mind if $\cot 2>\cot 3$ or $\cot 3>\cot 2.$

 

[Ray Vickson]

Homework Statement

Solution set of the inequality (cot^-1(x))² -(5 cot^-1(x)) +6 >0 is?

Homework Equations

The Attempt at a Solution

Subs cot^-1(x)=y
We get a quadratic inequality in y.
y²-5y+6>0
(y-2)(y-3)>0
Using the wavy curve method, the solution set is,
y∈(-∞,2) ∪(3,∞)
So cot^-1(x)<2 and cot^-1(x)>3
Taking cot on both sides of the inequality,
x<cot2 and x>cot3
x∈(-∞,cot2) ∪(cot3,∞)
Yet the answer is (-∞,cot3)∪(cot2,∞).
I'm guessing that in the step where I take cot on both sides, I'll have to change the inequality signs as arccot is a decreasing function. Is that where the problem lies?

What is the "wavy curve method"?

To clarify the answer you were given, plot the curve $y = \text{arccos}(x)$ over a broad range of $x$, such as $-10 \leq x \leq 10$ to see what the regions $\text{arccos}(x) < 2$ and $\text{arccos}(x) > 3$ look like along the $x$-axis.

 

[SammyS]

@takando12 ,

What is the range of the arccotangent function as you are using it in your course?

 

[ehild]

The plot of y=arccotan(x). In what interval of x is y≤2 or y≥3?