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January 25th, 2017, 02:03 PM   #1
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3^sin2x=2

This was a test question I don't know how to do it, if you can help me please!
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January 25th, 2017, 02:36 PM   #2
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$\displaystyle 3^{\sin2x}=2$
$\displaystyle 3^{\sin2x}=2=3^{\log_3 2}$ $\displaystyle \; \;$ $\displaystyle \;$ $\displaystyle \sin2x=\log_3 2$
$\displaystyle 2x=\arcsin (\log_3 2)$
$\displaystyle x=\frac{1}{2} \arcsin(\log_3 2)$
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Last edited by skipjack; January 25th, 2017 at 08:13 PM.
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January 25th, 2017, 02:42 PM   #3
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$3^{\sin(2x)}=2$

$\sin(2x) = \log_3(2)$

$\sin(\pi - 2x) = \log_3(2)$ as well
--------------------------------------------
$\sin(2x) = \log_3(2)$

$\sin(2x + 2k\pi) = \log_3(2),~k \in \mathbb{Z}$

$2(x+k \pi) = \arcsin(\log_3(2))$

$x = \dfrac{\arcsin(\log_3(2))}{2}-k \pi,~k \in \mathbb{Z}$

--------
$\sin(\pi - 2x) = \log_3(2)$

$\sin(\pi - 2x+ 2k\pi) = \log_3(2),~k \in \mathbb{Z}$

$\sin((2k+1) \pi - 2x ) = \log_3(2),~k \in \mathbb{Z}$

$(2k+1)\pi - 2x = \arcsin(\log_3(2))$

$-2x = \arcsin(\log_3(2))-(2k+1)\pi$

$x = \dfrac{(2k+1)\pi - \arcsin(\log_3(2))}{2}$

so the final result is

$x \in \left\{\dfrac{\arcsin(\log_3(2))}{2}-k \pi\right \} \cup \left\{\dfrac{(2k+1)\pi - \arcsin(\log_3(2))}{2}\right\},~k \in \mathbb{Z}$
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January 25th, 2017, 02:43 PM   #4
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I don't know what ark is.
I'm only in advance functions.
Maybe do you know another way, besides using ark?

Last edited by Abhikh; January 25th, 2017 at 02:54 PM.
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January 25th, 2017, 04:37 PM   #5
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Quote:
Originally Posted by Abhikh View Post
I don't know what ark is.
I'm only in advance functions.
Maybe do you know another way, besides using ark?
arcsin(x), asn(x), and $\displaystyle \sin^{-1}(x)$ are all different ways to say "inverse sine of x".

Frankly, I prefer to use asn(x) but that is a notation that isn't used much any more, so when I'm on the forums I use $\displaystyle \sin^{-1}(x)$.

-Dan

Last edited by skipjack; January 25th, 2017 at 08:15 PM.
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January 25th, 2017, 06:45 PM   #6
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I avoid $\sin^{-1}x$ as it easily creates confusion between $\arcsin x$ and $ \frac{1}{\sin x}$, especially if you wish to write $\sin^2 x = (\sin x)^2$.
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January 25th, 2017, 08:19 PM   #7
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Quote:
Originally Posted by Abhikh View Post
This was a test question . . .
Did you present it here in full, exactly as it appeared in the test?
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February 2nd, 2017, 01:55 PM   #8
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I believe that skipjack's point is that "3^sin(2x)= 2" is not a "question", it is an equation. What were you asked to do with that equation, solve it for x?
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