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 April 21st, 2019, 04:14 AM #1 Newbie   Joined: Apr 2019 From: Russia Posts: 3 Thanks: 0 Trigonometric equations I am preparing for my math exam and I need your help solving these trigonometric equations: $\sin^6x + \cos^6x = \frac{7}{16}$ $\sin^2 x + \sin^2 2x = 1$ $\sin 3x + \sin 2x = \sin x$ Last edited by greg1313; April 21st, 2019 at 04:36 AM. April 21st, 2019, 05:24 AM #2 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,978 Thanks: 1160 Math Focus: Elementary mathematics and beyond $\displaystyle \sin^6(x)+\cos^6(x)=\frac{7}{16}$ $\displaystyle (\sin^2(x)+\cos^2(x))^3-3\sin^4(x)\cos^2(x)-3\sin^2(x)\cos^4(x)=\frac{7}{16}$ $\displaystyle 1-3\sin^4(x)\cos^2(x)-3\sin^2(x)\cos^4(x)=\frac{7}{16}$ $\displaystyle 3\sin^2(x)\cos^2(x)(\sin^2(x)+\cos^2(x))=\frac{9}{ 16}$ $\displaystyle \sin^2(x)\cos^2(x)=\frac{3}{16}$ $\displaystyle \sin^2(2x)=\frac34$ $\displaystyle \sin(2x)=\pm\frac{\sqrt3}{2}$ I'll leave it to you to finish up. If you get stuck post back. Thanks from Maschke, topsquark and Celia8241 Last edited by greg1313; April 21st, 2019 at 10:29 AM. April 21st, 2019, 06:04 AM #3 Newbie   Joined: Apr 2019 From: Russia Posts: 3 Thanks: 0 Тrigonometric equation Okay, I solved it. Can you help me with others, please? Last edited by skipjack; April 21st, 2019 at 07:13 AM. April 21st, 2019, 07:35 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,116 Thanks: 2331 For the second equation, use the identities $\sin^2(x) = 1 - \cos^2(x)$ and $\sin(2x) = 2\sin(x)\cos(x)$. For the third equation, use the identity $\sin(3x) - \sin(x) = 2\cos(2x)\sin(x)$ and then the identity $\cos(2x) + \cos(x) = 2\cos(3x/2)\cos(x/2)$. Thanks from topsquark April 21st, 2019, 10:40 AM #5 Newbie   Joined: Apr 2019 From: Russia Posts: 3 Thanks: 0 Trigonometric equations " $\displaystyle \sin^2(x)\cos^2(x)=\frac{3}{16}$ $\displaystyle \sin^2(2x)=\frac34$ " Can someone explain it for me, please. I don't understand it. Last edited by Celia8241; April 21st, 2019 at 10:47 AM. April 21st, 2019, 10:55 AM   #6
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Quote:
 Originally Posted by Celia8241 " $\displaystyle \sin^2(x)\cos^2(x)=\frac{3}{16}$ $\displaystyle \sin^2(2x)=\frac34$ " Can someone explain it for me, please. I don't understand it.
starting with the basic double angle identity for sine ...

$2\sin{x}\cos{x} = \sin(2x) \implies 4\sin^2{x}\cos^2{x} = \sin^2(2x) \implies \sin^2{x}\cos^2{x} = \dfrac{\sin^2(2x)}{4} = \dfrac{3}{16} \implies \sin^2(2x) = \dfrac{12}{16} = \dfrac{3}{4}$ April 21st, 2019, 03:26 PM   #7
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Quote:
 Originally Posted by Celia8241 $\sin 3x + \sin 2x = \sin x$
$\displaystyle \sin(3x)=\sin(2x+x)=\sin(2x)\cos(x)+\sin(x)\cos(2x )=2\sin(x)\cos^2(x)+\sin(x)\cos(2x)$

$\displaystyle 2\sin(x)\cos^2(x)+\sin(x)\cos(2x)+2\sin(x)\cos(x)= \sin(x)$

Note that $\sin(x)=0$ is a solution. If $\sin(x)\ne0$ then,

$\displaystyle 2\cos^2(x)+\cos(2x)+2\cos(x)=1$

$\displaystyle 2\cos^2(x)+2\cos^2(x)-1+2\cos(x)-1=0$

$\displaystyle 4\cos^2(x)+2\cos(x)-2=0$

$\displaystyle (2\cos(x)-1)(2\cos(x)+2)=0$

Again, I'll leave it to you to finish up. April 22nd, 2019, 05:05 PM   #8
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Quote:
 Originally Posted by greg1313 $\displaystyle 2\cos^2(x)+\cos(2x)+2\cos(x)=1$
This equation is equivalent to $\cos(2x) = -\cos(x) = \cos(\pi - x)$,
so $\pm2x + 2k\pi = \pi - x$, where $k$ is an integer, etc. Tags equations, trigonometric Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post bilano99 Trigonometry 2 November 20th, 2012 08:49 AM bilano99 Trigonometry 2 November 16th, 2012 01:08 PM pksin Trigonometry 1 November 14th, 2012 02:07 AM filipd Trigonometry 8 November 3rd, 2011 12:36 AM symmetry Algebra 5 July 2nd, 2007 12:18 PM

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