Question

solve cotX^2+4cotX-7=0

Answer 1

find the general solution

Answer 2

@Michele_Laino

Answer 3

you can at first say,

let \(\cot(x)=a\), which will give you the following quadratic equation,

\(a^2+4a-7=0\)

Answer 4

Solve this equation for a, and after you find the solutions for a, you can find the value of x, using the fact that \(a=\cot(x)\)

Answer 5

(well, not the value of x, but a set of solutions, but I hope you see what I mean)

Answer 6

what about the general solution?

Answer 7

When you write \(\cot x^2\), do you really mean \(\cot^2 x\) ?

Answer 8

the latter

Answer 9

yes, mathstudent55 I would assume so:)
I didn't even notice that mistake (if it is a mistake)

Answer 10

have you solved for "a" (using a substitution of a=cot(x) that I offered) ?

Answer 11

yes but that just gave me -2 +or-sqrt(44)

Answer 12

i got mixed up for some reason

Answer 13

a²-4a-7=0
a²-4a=7
a²-4a+4=7+4
a²-4a+4=11
(a-2)²=11
a-2=±√11
a=-2±√11

that is what I get

Answer 14

its +4a

Answer 15

oh, tnx for catching me

Answer 16

(y)

Answer 17

oh, my fault, i did the step incorrectly

Answer 18

a²-4a-7=0
a²-4a=7
a²-4a+4=7+4
a²-4a+4=11
(a-2)²=11
a-2=±√11
a=2±√11

that is what I get if -4a
------------------
thus with +4a
i would get

now with +4a i get

a²+4a-7=0
a²+4a=7
a²+4a+4=7+4
a²+4a+4=11
(a+2)²=11
a+2=±√11
a=-2±√11

Answer 19

and by the quadratic formula
a²+4a-7=0

\(\large\color{black}{ \displaystyle \frac{-4\pm\sqrt{(4)^2-4(-7)(1)} }{2} }\)

\(\large\color{black}{ \displaystyle \frac{-4\pm\sqrt{44} }{2} }\)

\(\large\color{black}{ \displaystyle \frac{-4}{2}\pm\frac{\sqrt{44} }{2} }\)

Answer 20

see I am getting the same thing
a = -2±√11

Answer 21

okay, yay thanks, i get it now!

Answer 22

yes, so

\(\cot(x)=-2+\sqrt{11}\)
or
\(\cot(x)=-2-\sqrt{11}\)

Answer 23

\(\cot^{-1}(-2+\sqrt{11})=a\)

or

\(\cot^{-1}(-2-\sqrt{11})=a\)