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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: 6,766 Thanks: 698  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 
Newbie Joined: Mar 2014 Posts: 5 Thanks: 0  Re: solve the following equation  Hard problem
Here is the math problem

March 4th, 2014, 11:47 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,747 Thanks: 2133 
That equation has no real solutions. Are you sure it is given correctly?

March 4th, 2014, 01:40 PM  #6 
Newbie Joined: Mar 2014 Posts: 5 Thanks: 0  Re: solve the following equation  Hard problem
I double check on the problem and Yes!

March 4th, 2014, 04:18 PM  #7 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,590 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: 20,747 Thanks: 2133 
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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