Question

Question time

Answer 1

What are all the values of "x" between 0 and 2 pi that satisfy the equation?

\[(5+2\sqrt{6})^{\sin x} +(5-2\sqrt{6})^{\sin x} = 2 \sqrt{3}\]

Answer 2

oh wow

Answer 3

my exact thoughts. I have never seen anything like this

Answer 4

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

Answer 5

Okay I think maybe I got it!

Answer 6

Hmm

\[(5+2\sqrt{6})^{\sin x} +(\frac{1}{5 + 2\sqrt{6}})^{\sin x} = 2 \sqrt{3}\]

How about this? :-D

Answer 7

o...

Answer 8

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

Answer 9

I think I got it

Answer 10

awesome ishaan!

Answer 11

Let (5 + 2 sqrt (6)) be a

Let sinx be b

\[a^b + \frac{1}{a^b} = 2 \sqrt{3}\]

\[\frac{a^{2b} + 1}{a^b} = 2 \sqrt{3}\]

\[a^{2b} + 1 - 2 a^b \sqrt{3}\]

Answer 12

Let a^b be c

\[c^2 -2c \sqrt{3} + 1 = 0 \]

Answer 13

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

Answer 14

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

Answer 15

Thanks for the hint :3

Answer 16

No Problem :-)

Answer 17

x = 60, 150, 210, 330 degrees

Answer 18

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

Answer 19

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

Answer 20

is someone answer this limit question please

Answer 21

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

Answer 22

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

Answer 23

0
2
ali.athar 0
limx→1 x3−1/x2−1

Answer 24

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

Answer 25

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