M mathwhat Apr 2010 16 1 May 21, 2010 #1 can someone please help me to find the x: \(\displaystyle 2\cos(2x)-2\sin(x)=0\) another one: \(\displaystyle 2\sin(x)+\sin(2x)=0\) Thank you very much (Hi) Last edited: May 21, 2010

can someone please help me to find the x: \(\displaystyle 2\cos(2x)-2\sin(x)=0\) another one: \(\displaystyle 2\sin(x)+\sin(2x)=0\) Thank you very much (Hi)

M mathwhat Apr 2010 16 1 May 21, 2010 #3 Thank you very much but i did it exactly like you did and in the book the answers are different... for 1: \(\displaystyle \frac{\pi}{6}\) \(\displaystyle \frac{5}{6}\pi\) and for 2: \(\displaystyle \pi\) \(\displaystyle 2\pi\)

Thank you very much but i did it exactly like you did and in the book the answers are different... for 1: \(\displaystyle \frac{\pi}{6}\) \(\displaystyle \frac{5}{6}\pi\) and for 2: \(\displaystyle \pi\) \(\displaystyle 2\pi\)

harish21 Feb 2010 1,036 386 Dirty South May 21, 2010 #4 mathwhat said: Thank you very much but i did it exactly like you did and in the book the answers are different... for 1: \(\displaystyle \frac{\pi}{6}\) \(\displaystyle \frac{5}{6}\pi\) and for 2: \(\displaystyle \pi\) \(\displaystyle 2\pi\) Click to expand... The answers you have stated here are correct for the interval from \(\displaystyle 0\) to \(\displaystyle 2\pi\)

mathwhat said: Thank you very much but i did it exactly like you did and in the book the answers are different... for 1: \(\displaystyle \frac{\pi}{6}\) \(\displaystyle \frac{5}{6}\pi\) and for 2: \(\displaystyle \pi\) \(\displaystyle 2\pi\) Click to expand... The answers you have stated here are correct for the interval from \(\displaystyle 0\) to \(\displaystyle 2\pi\)

mr fantastic MHF Hall of Fame Dec 2007 16,948 6,768 Zeitgeist May 22, 2010 #5 dhiab said: Click to expand... Small error: \(\displaystyle \cos (2x) = 1 - 2\sin^2 (x)\) NOT \(\displaystyle 2\sin^2 (x) - 1\). So for #1 you will get pi/6 and 5pi/6 ....

dhiab said: Click to expand... Small error: \(\displaystyle \cos (2x) = 1 - 2\sin^2 (x)\) NOT \(\displaystyle 2\sin^2 (x) - 1\). So for #1 you will get pi/6 and 5pi/6 ....