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 January 25th, 2017, 01:03 PM #1 Newbie   Joined: Jan 2017 From: Canada Posts: 2 Thanks: 0 3^sin2x=2 This was a test question I don't know how to do it, if you can help me please!
 January 25th, 2017, 01:36 PM #2 Senior Member   Joined: Dec 2015 From: Earth Posts: 248 Thanks: 27 $\displaystyle 3^{\sin2x}=2$ $\displaystyle 3^{\sin2x}=2=3^{\log_3 2}$ $\displaystyle \; \;$ $\displaystyle \;$ $\displaystyle \sin2x=\log_3 2$ $\displaystyle 2x=\arcsin (\log_3 2)$ $\displaystyle x=\frac{1}{2} \arcsin(\log_3 2)$ Thanks from topsquark Last edited by skipjack; January 25th, 2017 at 07:13 PM.
 January 25th, 2017, 01:42 PM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 2,122 Thanks: 1102 $3^{\sin(2x)}=2$ $\sin(2x) = \log_3(2)$ $\sin(\pi - 2x) = \log_3(2)$ 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 $x \in \left\{\dfrac{\arcsin(\log_3(2))}{2}-k \pi\right \} \cup \left\{\dfrac{(2k+1)\pi - \arcsin(\log_3(2))}{2}\right\},~k \in \mathbb{Z}$ Thanks from topsquark
 January 25th, 2017, 01:43 PM #4 Newbie   Joined: Jan 2017 From: Canada Posts: 2 Thanks: 0 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 01:54 PM.
January 25th, 2017, 03:37 PM   #5
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Quote:
 Originally Posted by Abhikh 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 07:15 PM.

 January 25th, 2017, 05:45 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 Math Focus: Mainly analysis and algebra I avoid $\sin^{-1}x$ as it easily creates confusion between $\arcsin x$ and $\frac{1}{\sin x}$, especially if you wish to write $\sin^2 x = (\sin x)^2$. Thanks from topsquark
January 25th, 2017, 07:19 PM   #7
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Quote:
 Originally Posted by Abhikh This was a test question . . .
Did you present it here in full, exactly as it appeared in the test?

 February 2nd, 2017, 12:55 PM #8 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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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