anonymous
  • anonymous
Question time
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
What are all the values of "x" between 0 and 2 pi that satisfy the equation? \[(5+2\sqrt{6})^{\sin x} +(5-2\sqrt{6})^{\sin x} = 2 \sqrt{3}\]
anonymous
  • anonymous
oh wow
anonymous
  • anonymous
my exact thoughts. I have never seen anything like this

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More answers

anonymous
  • anonymous
Hmm \[(5 + 2 \sqrt6)(5 - 2\sqrt6) = 25 - 24 =1\]
anonymous
  • anonymous
Okay I think maybe I got it!
anonymous
  • anonymous
Hmm \[(5+2\sqrt{6})^{\sin x} +(\frac{1}{5 + 2\sqrt{6}})^{\sin x} = 2 \sqrt{3}\] How about this? :-D
anonymous
  • anonymous
o...
anonymous
  • anonymous
\[\frac{(5 + 2\sqrt{6})^{2\sin{x}} +1}{(5 + 2\sqrt{6})^{\sin{x}}} = 2\sqrt{3}\] Are we getting somewhere? lol
anonymous
  • anonymous
I think I got it
anonymous
  • anonymous
awesome ishaan!
anonymous
  • anonymous
Let (5 + 2 sqrt (6)) be a Let sinx be b \[a^b + \frac{1}{a^b} = 2 \sqrt{3}\] \[\frac{a^{2b} + 1}{a^b} = 2 \sqrt{3}\] \[a^{2b} + 1 - 2 a^b \sqrt{3}\]
anonymous
  • anonymous
Let a^b be c \[c^2 -2c \sqrt{3} + 1 = 0 \]
anonymous
  • 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....
anonymous
  • anonymous
Oh i see what you did.... Hmm nice
anonymous
  • anonymous
Thanks for the hint :3
anonymous
  • anonymous
No Problem :-)
anonymous
  • anonymous
x = 60, 150, 210, 330 degrees
anonymous
  • anonymous
Cool! :-D If only there was an easier way to get this
anonymous
  • anonymous
\[\lim_{x \rightarrow 1} x ^{3}-1/x ^{2}-1\]
anonymous
  • anonymous
is someone answer this limit question please
anonymous
  • anonymous
2 ali.athar 0 limx→1x3−1/x2−1 someone solve this question
anonymous
  • anonymous
From Mathematica:\[\text{Limit}\left[x^3-\frac{1}{x^2}-1,x\to 1\right]=-1 \]
anonymous
  • anonymous
0 2 ali.athar 0 limx→1 x3−1/x2−1
anonymous
  • anonymous
\[\lim_{x \rightarrow 2} x ^{3}-8/x ^{2}+x ^{}-6\]
anonymous
  • 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

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