Respuesta :
[tex]\bf sin(4x)=0\qquad [0,2\pi )\\\\
-----------------------------\\\\
\textit{Double Angle Identities}
\\ \quad \\
sin(2\theta)=2sin(\theta)cos(\theta)
\qquad \qquad
cos(2\theta)=
\begin{cases}
cos^2(\theta)-sin^2(\theta)\\
1-2sin^2(\theta)\\
2cos^2(\theta)-1
\end{cases}\\\\
-----------------------------\\\\
thus
\\\\\\
sin(4x)=0\implies sin[2(2x)]=0\implies 2sin(2x)cos(2x)=0
\\\\\\
2\left[ \boxed{2sin(x)cos(x)} \right]\left[\boxed{1-2sin^2(x)} \right]=0
\\\\\\
[/tex]
[tex]\bf \begin{cases} 2[2sin(x)cos(x)]=0\\\\ sin(x)=0\to x=sin^{-1}(0)\\ --------------\\ 0\qquad \pi \\ --------------\\ cos(x)=0\to x=cos^{-1}(0)\\ --------------\\ \frac{\pi }{2}\qquad \frac{3\pi }{2}\\ --------------\\ 1-2sin^2(x)=0\to 1=2sin^2(x)\\\\ \sqrt{\frac{1}{2}}=sin(x)\to \frac{1}{\sqrt{2}}=sin(x)\\\\ \frac{\sqrt{2}}{2}=sin(x)\to sin^{-1}\left( \frac{\sqrt{2}}{2} \right)=x\\ --------------\\ \frac{\pi }{4}\qquad \frac{3\pi }{4}\qquad \frac{5\pi }{4}\qquad \frac{7\pi }{4} \end{cases}[/tex]
[tex]\bf \begin{cases} 2[2sin(x)cos(x)]=0\\\\ sin(x)=0\to x=sin^{-1}(0)\\ --------------\\ 0\qquad \pi \\ --------------\\ cos(x)=0\to x=cos^{-1}(0)\\ --------------\\ \frac{\pi }{2}\qquad \frac{3\pi }{2}\\ --------------\\ 1-2sin^2(x)=0\to 1=2sin^2(x)\\\\ \sqrt{\frac{1}{2}}=sin(x)\to \frac{1}{\sqrt{2}}=sin(x)\\\\ \frac{\sqrt{2}}{2}=sin(x)\to sin^{-1}\left( \frac{\sqrt{2}}{2} \right)=x\\ --------------\\ \frac{\pi }{4}\qquad \frac{3\pi }{4}\qquad \frac{5\pi }{4}\qquad \frac{7\pi }{4} \end{cases}[/tex]
