Mathematics, 04.07.2019 07:30, 99keevintaylor012
10 points pls . find all general solutions: 2cos^2(3) − 1 = −sin(3)
Answers: 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)](/tpl/images/0049/3937/282b4.png)
![\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}](/tpl/images/0049/3937/50e52.png)
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10 points pls . find all general solutions: 2cos^2(3) − 1 = −sin(3)...
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