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November 16th, 2016, 01:16 PM   #1
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Trigonometrikus Pit

4 sin 2x + 4 cos 2x = 4
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November 16th, 2016, 03:29 PM   #2
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Quote:
Originally Posted by romsek View Post
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.
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November 16th, 2016, 03:36 PM   #3
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$$\begin{align*}
\tfrac1{\sqrt2}\sin(u) + \tfrac1{\sqrt2}\cos(u) &= \tfrac1{\sqrt2} \\
\sin{(\tfrac{\pi}4)}\sin{(u)} + \cos{(\tfrac{\pi}4)}\cos{(u)} &= \tfrac1{\sqrt2} \\
\cos{(u - \tfrac{\pi}4)} &= \tfrac1{\sqrt2} \\
u - \tfrac{\pi}4 &= 2n\pi \pm \tfrac{\pi}4
\end{align*}$$
... and so on.
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November 17th, 2016, 09:09 AM   #4
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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 09:22 AM.
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November 24th, 2016, 08:40 AM   #5
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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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