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 April 21st, 2019, 03:14 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 03:36 AM.
 April 21st, 2019, 04:24 AM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,934 Thanks: 1128 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 09:29 AM.
 April 21st, 2019, 05:04 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 06:13 AM.
 April 21st, 2019, 06:35 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,623 Thanks: 2076 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, 09:40 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 09:47 AM.
April 21st, 2019, 09:55 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, 02:26 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, 04:05 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.

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