## anonymous 5 years ago Question time

1. anonymous

What are all the values of "x" between 0 and 2 pi that satisfy the equation? $(5+2\sqrt{6})^{\sin x} +(5-2\sqrt{6})^{\sin x} = 2 \sqrt{3}$

2. anonymous

oh wow

3. anonymous

my exact thoughts. I have never seen anything like this

4. anonymous

Hmm $(5 + 2 \sqrt6)(5 - 2\sqrt6) = 25 - 24 =1$

5. anonymous

Okay I think maybe I got it!

6. anonymous

Hmm $(5+2\sqrt{6})^{\sin x} +(\frac{1}{5 + 2\sqrt{6}})^{\sin x} = 2 \sqrt{3}$ How about this? :-D

7. anonymous

o...

8. anonymous

$\frac{(5 + 2\sqrt{6})^{2\sin{x}} +1}{(5 + 2\sqrt{6})^{\sin{x}}} = 2\sqrt{3}$ Are we getting somewhere? lol

9. anonymous

I think I got it

10. anonymous

awesome ishaan!

11. anonymous

Let (5 + 2 sqrt (6)) be a Let sinx be b $a^b + \frac{1}{a^b} = 2 \sqrt{3}$ $\frac{a^{2b} + 1}{a^b} = 2 \sqrt{3}$ $a^{2b} + 1 - 2 a^b \sqrt{3}$

12. anonymous

Let a^b be c $c^2 -2c \sqrt{3} + 1 = 0$

13. anonymous

$(5 + 2\sqrt{6})^{2\sin{x}} + 1 = 2\sqrt{3}\times (5 + 2\sqrt{6})^{\sin{x}}$ $1 = 2\sqrt{3}\times (5 + 2\sqrt{6})^{\sin{x}} - (5 + 2\sqrt{6})^{2\sin{x}}$ Hmm....

14. anonymous

Oh i see what you did.... Hmm nice

15. anonymous

Thanks for the hint :3

16. anonymous

No Problem :-)

17. anonymous

x = 60, 150, 210, 330 degrees

18. anonymous

Cool! :-D If only there was an easier way to get this

19. anonymous

$\lim_{x \rightarrow 1} x ^{3}-1/x ^{2}-1$

20. anonymous

is someone answer this limit question please

21. anonymous

2 ali.athar 0 limx→1x3−1/x2−1 someone solve this question

22. anonymous

From Mathematica:$\text{Limit}\left[x^3-\frac{1}{x^2}-1,x\to 1\right]=-1$

23. anonymous

0 2 ali.athar 0 limx→1 x3−1/x2−1

24. anonymous

$\lim_{x \rightarrow 2} x ^{3}-8/x ^{2}+x ^{}-6$

25. anonymous

x^3 - 8 = (x-2) (x^2 + 2x + 4) x^2 + x - 6 = (x+3)(x-2) x-2 cancels out (x^2 + 2x + 4)/(x+3) = (4 + 4 + 4 )/ 5 = 12/5