Question

Find the solutions to the given equation on the interval [0,2π)?
cot^3x−cotx=0

I don't understand this...angles to recognize etc..
please explain..thank you

Answer 1

cot^3x−cotx=0
cotx(cot^2 x - 1) = 0
cotx(cotx+1)(cotx-1) = 0
so cotx = 0 , cotx = -1, cotx = 1

consider the last two:
if cotx = -1, cotx = 1
tanx = -1, tanx = 1
which means that x could be in any of the 4 quadrants
you must know that tan π/4 = 1 , (tan 45° =1)
so x = π/4 , 3π/4, 5π/4, and 7π/4

let's look at the cotx = 0
mathematically, then tanx = 1/0, which of course is undefined
which means that the tangent curve must have an asymptote at that value of x
looking at the tangent curve , we can see that at x = π/2 , or 90°, the tangent is undefined
The period of the tangent curve is π, so we have another solution at
π/2 + π = 3π/2

so you have 6 solutions for your equation:
x = π/2, 3π/2, π/4 , 3π/4, 5π/4, 7π/4

You should memorize the basic trig ratios for angles of
30°, 45°, 60° and 90° from the 2 simple right-angled triangles with sides
of 1-1-√2, and 3-4-5

Answer 2

Thank you so,so much for that explanation!