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 December 19th, 2018, 09:38 AM #1 Newbie   Joined: Dec 2018 From: Usa Posts: 2 Thanks: 0 Solving in R Hello, I recently came here invited by a friend of mine into the forum, bless who can give me as much as details about this ^^ 1) Solve in R cos(3x)=1/2 2) Calculate cos(3x) according to cos(x) Last edited by skipjack; December 19th, 2018 at 01:04 PM. December 19th, 2018, 09:47 AM   #2
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Quote:
 Originally Posted by AnasMathZ 1) Solve in R cos(3x)=1/2 2) Calculate cos(3x) according to cos(x)
note ...

$\cos(u) = \dfrac{1}{2}$ at $u = 2k\pi \pm \dfrac{\pi}{3} \, ; \, k \in \mathbb{Z}$ December 19th, 2018, 09:55 AM   #3
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 Originally Posted by skeeter note ... $\cos(u) = \dfrac{1}{2}$ at $u = 2k\pi \pm \dfrac{\pi}{3} \, ; \, k \in \mathbb{Z}$
you think you can solve it for me ? December 19th, 2018, 10:16 AM   #4
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 Originally Posted by AnasMathZ you think you can solve it for me ?
you try and solve it ... start by letting $u=3x$ December 19th, 2018, 12:50 PM   #5
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Quote:
 Originally Posted by AnasMathZ 1) Solve in R cos(3x)=1/2 2) Calculate cos(3x) according to cos(x)
skeeter has given you enough to solve 1). If you write $u=3x$ you get $\cos{(3x)} = \frac12$. You should know what values of $u$ give that result (and skeeter has told you that too). Then $3x=u$ gives you a simple equation to determine $x$.

For 2) you will need to use the identities \begin{align}\cos{(A+B)} &= \cos{(A)}\cos{(B)} - \sin{(A)}\sin{(B)} \\ \sin{(A+B)} &= \sin{(A)}\cos{(B)} + \cos{(A)}\sin{(B)} &\text{and} \\ \cos^2{(A)} + \sin^2{(A)} &= 1\end{align}
to turn $(3x)$ into $(2x + x)$, $(2x)$ into $(x + x)$ and finally eliminate $\sin^2{(x)}$ from the result. Tags solving 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 Rakov12 Algebra 3 April 29th, 2018 03:28 PM Mircode Real Analysis 0 February 4th, 2013 12:22 PM JohnC Algebra 6 April 17th, 2011 10:06 PM TimNice Linear Algebra 2 April 11th, 2010 12:04 PM peekay Calculus 1 March 13th, 2009 01:48 PM

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