November 16th, 2016, 01:16 PM  #1 
Newbie Joined: Nov 2016 From: Hungary Posts: 3 Thanks: 0  Trigonometrikus Pit
4 sin 2x + 4 cos 2x = 4

November 16th, 2016, 03:29 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,032 Thanks: 2342 Math Focus: Mainly analysis and algebra  $$\begin{align*} \tfrac1{\sqrt2}\sin(u) + \tfrac1{\sqrt2}\cos(u) &= \tfrac1{\sqrt2} \\ \cos{(\tfrac{\pi}4)}\sin{(u)} + \sin{(\tfrac{\pi}4)}\cos{(u)} &= \tfrac1{\sqrt2} \\ \sin{(u + \tfrac{\pi}4)} &= \tfrac1{\sqrt2} \\ u + \tfrac{\pi}4 &= (4n + 1)\tfrac{\pi}2 \pm \tfrac{\pi}4 \end{align*}$$ ... and so on. 
November 16th, 2016, 03:36 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,032 Thanks: 2342 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. 
November 17th, 2016, 09:09 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,822 Thanks: 750 
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. 
November 24th, 2016, 08:40 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 