Find the x

Apr 2010
16
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:
Apr 2010
16
1
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:
\(\displaystyle \pi\)

\(\displaystyle 2\pi\)
 
Feb 2010
1,036
386
Dirty South
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:
\(\displaystyle \pi\)

\(\displaystyle 2\pi\)
The answers you have stated here are correct for the interval from \(\displaystyle 0\) to \(\displaystyle 2\pi\)
 

mr fantastic

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Dec 2007
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Zeitgeist
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 ....