sin(4x) = sqrt(3)/2: solve on interval [0, 2pi]

[Striker]

i need help with a problem involving sin(4x) i know the double angle identities but have no idea what to do here.

the equation needs to be solved on the interval from 0 to 2pi

sin(4x)= sqrt(3)/2

thanks for any help you can give me

 

[pka]

One does not need double-angles to work this one.
If \(\displaystyle \sin (A) = \frac{{\sqrt 3 }}{2}\) the we know that \(\displaystyle A = \frac{\pi }{3}\quad or\quad A = \frac{{2\pi }}{3}\).

Thus \(\displaystyle 4x = \frac{\pi }{3}\quad or\quad 4x = \frac{{2\pi }}{3}\).

Solve for x.
\(\displaystyle \begin{array}{l}
x = \frac{\pi }{{12}} + \frac{{\pi k}}{2},\quad k = 0,1,2,3 \\
x = \frac{\pi }{6} + \frac{{\pi k}}{2},\quad k = 0,1,2,3 \\
\end{array}\)

 

[soroban]

Re: sin(4x)

Hello, Striker!

No identities are needed . . .

Solve for \(\displaystyle \,0\,\leq\,x\,\leq\,2\pi\)

. . \(\displaystyle \L\sin(4x) \;= \;\frac{\sqrt{3}}{2}\)

We have: \(\displaystyle \L\:4x \;=\;\frac{\pi}{3},\:\frac{2\pi}{3},\:\frac{7\pi}{3},\:\frac{8\pi}{3},\:\frac{13\pi}{3},\:\frac{14\pi}{3},\:\frac{19\pi}{3},\:\frac{20\pi}{3}\)

Therefore: \(\displaystyle \L\:x\;=\;\frac{\pi}{12},\:\frac{\pi}{6},\:\frac{7\pi}{12},\:\frac{2\pi}{3},\:\frac{13\pi}{12},\:\frac{7\pi}{6},\:\frac{19\pi}{12},\:\frac{5\pi}{3}\)