Mathematics, 03.02.2020 07:54 rociomartinez9
Sin4x=0. find all the solutions in the interval [0, 2pi)
Answers: 2
![\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 \\\\\\ ](/tpl/images/0495/7268/40c7e.png)
![\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}](/tpl/images/0495/7268/8e273.png)
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Sin4x=0. find all the solutions in the interval [0, 2pi)...
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