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
Last edited by mathwhat; May 21st 2010 at 07:55 AM.
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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: $\displaystyle \pi$ $\displaystyle 2\pi$
Originally Posted by mathwhat 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 $\displaystyle 2\pi$
Originally Posted by dhiab 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 ....