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 April 12th, 2012, 02:17 AM #1 Newbie   Joined: Apr 2012 Posts: 2 Thanks: 0 How many solutions to an equation Hi folks, I am trying to help my son with his Maths revision but too many years and brain cells have been and gone. Can someone help to explain how we should approach this problem? How many solutions does the equation (4sinx - 5root5)(sin x+1)=0 have in the interval 0?x<2? Thanks
April 12th, 2012, 04:19 AM   #2
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Re: How many solutions to an equation

Hello, raycol1970!

Quote:
 $\text{How many solutions does the equation }\,(4\,\!\sin x\,-\,5\sqrt{5})(\sin x\,+\,1)\:=\:0\,\text{ have} \;\;\;\;\text{in the interval }0\,\le\,x\,<\,2\pi$

$\begin{Bmatrix}4\,\!\sin x\,-\,5\sqrt{5} \:=\:0 &\;\;\Rightarrow\;\;=&\sin x \:=\:\frac{5\sqrt{5}}{4} &\;\;\Rightarrow\;\;=&\text{not real} \\ \\ \\ \sin x\,+\,1\:=\:0 &\Rightarrow=&\sin x \:=\:-1 &\Rightarrow=&x \:=\:\frac{3\pi}{2} \end{Bmatrix}$

$\text{Answer: }\,\text{one solution.}$

 April 13th, 2012, 04:19 AM #3 Newbie   Joined: Apr 2012 Posts: 2 Thanks: 0 Re: How many solutions to an equation Soraban - thank you ever so much! - listen really sorry I made a mistake in the question it should be just root5 and not 5root5 - really sorry for this...
April 14th, 2012, 02:03 PM   #4
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Re: How many solutions to an equation

Quote:
 Originally Posted by raycol1970 Soraban - thank you ever so much! - listen really sorry I made a mistake in the question it should be just root5 and not 5root5 - really sorry for this...
You have only to replace $5\sqrt{5}$ with $\sqrt{5}$ in soroban's solution. But $\frac{\sqrt{5}}{4}<1$ so $\sin x= \frac{\sqrt{5}}{4}$ has a real solution.

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