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April 21st, 2019, 03:14 AM   #1
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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.
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April 21st, 2019, 04:24 AM   #2
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$\displaystyle \sin^6(x)+\cos^6(x)=\frac{7}{16}$

$\displaystyle (\sin^2(x)+\cos^2(x))^3-3\sin^4(x)\cos^2(x)-3\sin^2(x)\cos^4(x)=\frac{7}{16}$

$\displaystyle 1-3\sin^4(x)\cos^2(x)-3\sin^2(x)\cos^4(x)=\frac{7}{16}$

$\displaystyle 3\sin^2(x)\cos^2(x)(\sin^2(x)+\cos^2(x))=\frac{9}{ 16}$

$\displaystyle \sin^2(x)\cos^2(x)=\frac{3}{16}$

$\displaystyle \sin^2(2x)=\frac34$

$\displaystyle \sin(2x)=\pm\frac{\sqrt3}{2}$

I'll leave it to you to finish up. If you get stuck post back.
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Last edited by greg1313; April 21st, 2019 at 09:29 AM.
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April 21st, 2019, 05:04 AM   #3
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Đ¢rigonometric equation

Okay, I solved it. Can you help me with others, please?

Last edited by skipjack; April 21st, 2019 at 06:13 AM.
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April 21st, 2019, 06:35 AM   #4
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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 $\cos(2x) + \cos(x) = 2\cos(3x/2)\cos(x/2)$.
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April 21st, 2019, 09:40 AM   #5
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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.
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April 21st, 2019, 09:55 AM   #6
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Quote:
Originally Posted by Celia8241 View Post
" $\displaystyle \sin^2(x)\cos^2(x)=\frac{3}{16}$

$\displaystyle \sin^2(2x)=\frac34$ "

Can someone explain it for me, please. I don't understand it.
starting with the basic double angle identity for sine ...

$2\sin{x}\cos{x} = \sin(2x) \implies 4\sin^2{x}\cos^2{x} = \sin^2(2x) \implies \sin^2{x}\cos^2{x} = \dfrac{\sin^2(2x)}{4} = \dfrac{3}{16} \implies \sin^2(2x) = \dfrac{12}{16} = \dfrac{3}{4}$
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April 21st, 2019, 02:26 PM   #7
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Quote:
Originally Posted by Celia8241 View Post
$\sin 3x + \sin 2x = \sin x$
$\displaystyle \sin(3x)=\sin(2x+x)=\sin(2x)\cos(x)+\sin(x)\cos(2x )=2\sin(x)\cos^2(x)+\sin(x)\cos(2x)$

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

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.
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April 22nd, 2019, 04:05 PM   #8
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Quote:
Originally Posted by greg1313 View Post
$\displaystyle 2\cos^2(x)+\cos(2x)+2\cos(x)=1$
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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