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 November 16th, 2016, 12:16 PM #1 Newbie   Joined: Nov 2016 From: Hungary Posts: 3 Thanks: 0 Trigonometrikus Pit 4 sin 2x + 4 cos 2x = 4
November 16th, 2016, 02:29 PM   #2
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Joined: Dec 2013
From: Colombia

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Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by romsek let $u=2x$ $\sin(u) + \cos(u) = 1$
\begin{align*} \tfrac1{\sqrt2}\sin(u) + \tfrac1{\sqrt2}\cos(u) &= \tfrac1{\sqrt2} \\ \cos{(\tfrac{\pi}4)}\sin{(u)} + \sin{(\tfrac{\pi}4)}\cos{(u)} &= \tfrac1{\sqrt2} \\ \sin{(u + \tfrac{\pi}4)} &= \tfrac1{\sqrt2} \\ u + \tfrac{\pi}4 &= (4n + 1)\tfrac{\pi}2 \pm \tfrac{\pi}4 \end{align*}
... and so on.

 November 16th, 2016, 02:36 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra \begin{align*} \tfrac1{\sqrt2}\sin(u) + \tfrac1{\sqrt2}\cos(u) &= \tfrac1{\sqrt2} \\ \sin{(\tfrac{\pi}4)}\sin{(u)} + \cos{(\tfrac{\pi}4)}\cos{(u)} &= \tfrac1{\sqrt2} \\ \cos{(u - \tfrac{\pi}4)} &= \tfrac1{\sqrt2} \\ u - \tfrac{\pi}4 &= 2n\pi \pm \tfrac{\pi}4 \end{align*} ... and so on. Thanks from topsquark and gab17
 November 17th, 2016, 08:09 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,199 Thanks: 873 If, as you say in another post, this is supposed to have been $\displaystyle 4\sin^2(x)+ 4\cos^2(x)= 4$ (4 sin^2(x) + 4 cos^2(x) = 4 if you don't want to use Latex) then, because $\displaystyle \sin^2(x)+ \cos^2(x)= 1$ for all x, that is true for all x. Last edited by skipjack; November 17th, 2016 at 08:22 AM.
 November 24th, 2016, 07:40 AM #5 Newbie   Joined: Nov 2016 From: Canada Posts: 13 Thanks: 3 Math Focus: All Solution is x = n*PI + PI/4, for all n belongs to Z and x = n*PI, for all n belongs to Z

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