Question

10 points pls help. Find all general solutions: 2cos^2(3) − 1 = −sin(3)

Answer 1

$\bf \textit{Double Angle Identities} \\\\ sin(2\theta)=2sin(\theta)cos(\theta) \\\\ cos(2\theta)= \begin{cases} cos^2(\theta)-sin^2(\theta)\\ 1-2sin^2(\theta)\\ 2cos^2(\theta)-1 \end{cases}\quad \begin{array}{llll} \\ \leftarrow \textit{we'll use}\\ \leftarrow \textit{these two} \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{double-angle identity}}{2cos^2(3x)-1}=-sin(3x)\implies cos[2(3x)]=-sin(3x)$

$\bf 1-2sin^2(3x)=-sin(3x)\implies 0=\stackrel{\stackrel{ax^2+bx+c}{\downarrow }}{2sin^2(3x)-sin(3x)-1} \\\\\\ 0=[2sin(3x)+1][sin(3x)-1] \\\\[-0.35em] ~\dotfill\\\\ 0=2sin(3x)+1\implies -1=2sin(3x)\implies -\cfrac{1}{2}=sin(3x) \\\\\\ \cfrac{7\pi }{6}~,~\cfrac{11\pi }{6}=3x\implies \boxed{\cfrac{7\pi }{18}~,~\cfrac{11\pi }{18}=x} \\\\[-0.35em] ~\dotfill\\\\ 0=sin(3x)-1\implies 1=sin(3x)\implies \cfrac{\pi }{2}=3x\implies \boxed{\cfrac{\pi }{6}=x}$

now, those angles are the angles in the range of [0, 2π] only.

a general solution angles will be

$\bf \measuredangle x=\stackrel{\textit{first case}}{\cfrac{7\pi }{18}+2\pi n~~,~~\cfrac{11\pi }{18}+2\pi n}~~,~~\stackrel{\textit{second case}}{\cfrac{\pi }{6}+2\pi n}\qquad \qquad \begin{array}{llll} where\\ n\in \mathbb{Z} \end{array}$