Solve each equation on the interval [0,2pi] 1) sin^2x-2cosx-2=0 2) 4cos^x=5-4sinx 3) sin2x=cosx 4) cos2x=cosx 5) cos2x+5cosx+3=0
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Originally Posted by LillyJackson06 Solve each equation on the interval [0,2pi] 1) sin^2x-2cosx-2=0 2) 4cos^x=5-4sinx 3) sin2x=cosx 4) cos2x=cosx 5) cos2x+5cosx+3=0 Please show what work you have been able to do. We are not a homework service. -Dan
I do not know how to start these equations, and what to substitute in, which is the first step, which i need help on.
Originally Posted by LillyJackson06 Solve each equation on the interval [0,2pi] 1) sin^2x-2cosx-2=0 2) 4cos^x=5-4sinx 3) sin2x=cosx 4) cos2x=cosx 5) cos2x+5cosx+3=0 Write #1 as $\displaystyle 1-\cos^2(x)-2\cos(x)-2=\cos^2(x)+2\cos(x)+1=0$. Then let $\displaystyle u=\cos(x)$ solve $\displaystyle u^2+2u+1=0$. You have left something out of #2. #3 $\displaystyle \sin(2x)=2\sin(x)\cos(x)$ #5 $\displaystyle \cos(2x)=2\cos^2(x)-1$
I now understand #1. Sorry, #2 should be 4cos^2x=5-4sinx. I also now get #3 and #5. Thanks so much.
Originally Posted by LillyJackson06 I now understand #1. Sorry, #2 should be 4cos^2x=5-4sinx. I also now get #3 and #5. #2 $\displaystyle 4(1-\sin^2(x))=5-4\sin(x)$
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