Question

Sin4x=0. Find all the solutions in the interval [0, 2pi)

Answer 1

`\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 \\\\\\`


`\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)\\ --------------\\ (\pi )/(2)\qquad (3\pi )/(2)\\ --------------\\ 1-2sin^2(x)=0\to 1=2sin^2(x)\\\\ \sqrt{(1)/(2)}=sin(x)\to (1)/(√(2))=sin(x)\\\\ (√(2))/(2)=sin(x)\to sin^(-1)\left( (√(2))/(2) \right)=x\\ --------------\\ (\pi )/(4)\qquad (3\pi )/(4)\qquad (5\pi )/(4)\qquad (7\pi )/(4) \end{cases}`