My Math Forum solve the following equation - Hard problem

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 March 2nd, 2014, 06:54 PM #1 Newbie   Joined: Mar 2014 Posts: 5 Thanks: 0 solve the following equation - Hard problem I need to show all the work for extra credit.
 March 3rd, 2014, 05:31 AM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: solve the following equation - Hard problem Your teacher said that if you showed someone else's work you would be given extra credit? That seems very strange.
 March 3rd, 2014, 12:23 PM #3 Global Moderator   Joined: May 2007 Posts: 6,821 Thanks: 723 Re: solve the following equation - Hard problem It is extremely difficult to solve an equation which is blank.
March 4th, 2014, 11:35 AM   #4
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Re: solve the following equation - Hard problem

Here is the math problem
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 March 4th, 2014, 11:47 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 That equation has no real solutions. Are you sure it is given correctly?
March 4th, 2014, 01:40 PM   #6
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Re: solve the following equation - Hard problem

I double check on the problem and Yes!
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 problem.JPG (10.0 KB, 114 views)

 March 4th, 2014, 04:18 PM #7 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: solve the following equation - Hard problem Well, you must mean "to the power"; then you get solutions: http://www.wolframalpha.com/input/?i=so ... %3D0+for+x The way you present your stuff, looks like multiplications.
 March 5th, 2014, 03:50 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2216 After "double checking", a different equation was posted, so the original equation was incorrect. It's easy to see that the second equation's real solutions correspond to sin²x = 1/4 and sin²x = 3/4, as the equation is effectively a quadratic equation in (16/81)^(sin²x), solving which gives (16/81)^(sin²x) = 2/3 or 8/27. Hence x = ±?/6 + k? or ±?/3 + k?, where k is an integer.

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