Question

2sin(4x)=1

Answer 1

\[\sin(4x)=1\]
\[4x=\pi/6+2*\pi*k\]
\[x=\frac{\pi}{24}+\frac{\pi*k}{2}\]

Answer 2

first line shoule be
2sin(4x)=1

Answer 3

thank you

Answer 4

let's wait for polpak - and see if I messed up :)

Answer 5

I'm pretty sure there's only 2 angles that correspond to positive 1/2

Answer 6

Sorry forgot to divide by 2n(pi) by 4 also.

Answer 7

\begin{array}{lrcl}
& 2\sin(4x) &= &1\\
\implies & sin(4x) & = &\frac{1}{2}\\
\implies & 4x & \in & \{\frac{\pi}{6} + 2n\pi, \frac{5\pi}{6}+ 2n\pi\}\\
\implies & x & \in &\{\frac{\pi}{24} + \frac{n\pi}{2}, \frac{5\pi}{24} + \frac{n\pi}{2}\}\\
\implies & x & \in &\{\frac{(12n + 1)\pi}{24}, \frac{(12n + 5)\pi}{24} \}\\
\end{array}

Answer 8

Pretty sure that's right.

Answer 9

yes that looks right to me. I was missing the second solution set (with the 5pi).