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 January 24th, 2012, 01:30 AM #1 Newbie   Joined: Jan 2012 Posts: 9 Thanks: 0 Trigonometric inequalities problem So I have problem solving this two trigonometric inequalities sinx+cox+sin2x>1 tg(x+pi/5)
January 24th, 2012, 02:31 AM   #2
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Re: Trigonometric inequalities problem

Quote:
 sinx+cox+sin2x>1
Is that 'cos' or 'cot'?

January 24th, 2012, 02:45 AM   #3
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Re: Trigonometric inequalities problem

Quote:
 tg(x+pi/5)
$\text{Using }\tan$$\frac{\pi}{3}$$\,=\,\sqrt{3}\text{ and }\tan$$\frac{4\pi}{3}$$\,=\,\sqrt{3}$

$\text{ the solutions on the interval }[0,\,2\pi]$

$\text{ can be determined by studying the unit circle and the associated trigonometric values.}$

 January 24th, 2012, 03:05 AM #4 Newbie   Joined: Jan 2012 Posts: 9 Thanks: 0 Re: Trigonometric inequalities problem oh,its cosx
 January 24th, 2012, 10:22 AM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Re: Trigonometric inequalities problem We are given: $\sin(x)+\cos(x)+\sin(2x)>1$ $\sin(x)+\cos(x)>\sin^2(x)-2\sin(x)\cos(x)+\cos^2(x)$ $\sin(x)+\cos(x)>$$\sin(x)-\cos(x)$$^2$ $\sqrt{2}\sin$$x+\frac{\pi}{4}$$>2\sin^2$$x-\frac{\pi}{4}$$$ $\sin$$x+\frac{\pi}{4}$$>\sqrt{2}$$1-\sin^2\(x+\frac{\pi}{4}$$\)$ $\sqrt{2}\sin^2$$x+\frac{\pi}{4}$$+\sin$$x+\frac{\p i}{4}$$-\sqrt{2}>0$ Solving the quadratic equation: $\sqrt{2}\sin^2$$x+\frac{\pi}{4}$$+\sin$$x+\frac{\p i}{4}$$-\sqrt{2}=0$ $\sin$$x+\frac{\pi}{4}$$=\frac{-1\pm3}{2\sqrt{2}}$ Assuming x is real, we must discard the root whose magnitude is greater than 1, so we have: $\sin$$x+\frac{\pi}{4}$$=\frac{-1+3}{2\sqrt{2}}=\frac{1}{\sqrt{2}}$ Checking the interval given by: $\frac{\pi}{4} $0 Using $x=\frac{\pi}{4}$, we find by substitution into the quadratic inequality: $\sqrt{2}\sin^2$$\frac{\pi}{4}+\frac{\pi}{4}$$+\sin $$\frac{\pi}{4}+\frac{\pi}{4}$$-\sqrt{2}>0$ $\sqrt{2}+1-\sqrt{2}>0$ $1>0$ Thus, we find: $2k\pi where $k\in\mathbb Z$
 January 24th, 2012, 03:27 PM #6 Newbie   Joined: Jan 2012 Posts: 9 Thanks: 0 Re: Trigonometric inequalities problem THANK YOU
January 24th, 2012, 05:44 PM   #7
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Re: Trigonometric inequalities problem

Quote:
 Originally Posted by niki500 tg(x+pi/5)
$\text{In the interval }[0,\,2\pi],$

$0\,\le\,x\,\le\frac{2\pi}{15}$

$\frac{3\pi}{10}\,\lt\,x\,\le\,\frac{17\pi}{15}$

$\frac{13\pi}{10}\,\lt\,x\,\le\,2\pi$

These values can be obtained by solving

$x\,+\,\frac{\pi}{5}\,=\,\frac{\pi}{3}$

$x\,+\,\frac{\pi}{5}\,=\,\frac{\pi}{2}$

$x\,+\,\frac{\pi}{5}\,=\,\frac{3\pi}{2}$

 January 25th, 2012, 01:06 AM #8 Newbie   Joined: Jan 2012 Posts: 9 Thanks: 0 Re: Trigonometric inequalities problem Ok,so I have these two more inequalities to solve and Im done 2sinx*sin2x*sin3x
 January 25th, 2012, 01:17 AM #9 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,879 Thanks: 1087 Math Focus: Elementary mathematics and beyond Re: Trigonometric inequalities problem Have you made any attempt at solving them?
 January 25th, 2012, 01:37 AM #10 Newbie   Joined: Jan 2012 Posts: 9 Thanks: 0 Re: Trigonometric inequalities problem Yes I did for the first one I come to this sin2xcos3xcosx>0 and for the second one sin12x>-sin4x and dont know how to solve it further

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