## anonymous one year ago solve cotX^2+4cotX-7=0

1. anonymous

find the general solution

2. anonymous

@Michele_Laino

3. SolomonZelman

you can at first say, let $$\cot(x)=a$$, which will give you the following quadratic equation, $$a^2+4a-7=0$$

4. SolomonZelman

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

5. SolomonZelman

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

6. anonymous

7. mathstudent55

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

8. anonymous

the latter

9. SolomonZelman

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

10. SolomonZelman

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

11. anonymous

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

12. SolomonZelman

i got mixed up for some reason

13. SolomonZelman

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

14. anonymous

its +4a

15. SolomonZelman

oh, tnx for catching me

16. anonymous

(y)

17. SolomonZelman

oh, my fault, i did the step incorrectly

18. SolomonZelman

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

19. SolomonZelman

and by the quadratic formula a²+4a-7=0 $$\large\color{black}{ \displaystyle \frac{-4\pm\sqrt{(4)^2-4(-7)(1)} }{2} }$$ $$\large\color{black}{ \displaystyle \frac{-4\pm\sqrt{44} }{2} }$$ $$\large\color{black}{ \displaystyle \frac{-4}{2}\pm\frac{\sqrt{44} }{2} }$$

20. SolomonZelman

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

21. anonymous

okay, yay thanks, i get it now!

22. SolomonZelman

yes, so $$\cot(x)=-2+\sqrt{11}$$ or $$\cot(x)=-2-\sqrt{11}$$

23. SolomonZelman

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