i!The solutions are x=pi/3,pi,(5pi)/3 Use this identity: cos2x=2cos^2x-1 Now, here's the actual problem: 2cos2x+2cosx=0 2(color(red)(cos2x))+2cosx=0 2(color(red)(2cos^2x-1))+2cosx=0 color(red)(4cos^2x-2)+2cosx=0 4cos^2x+2cosx-2=0 4(cosx)^2+2cosx-2=0 Substitute u for cosx: 4u^2+2u-2=0 2u^2+u-1=0 (2u-1)(u+1)=0 u=1/2,-1 Put cosx back in for u: cosx=1/2, qquadcosx=-1 x=pi/3,(5pi)/3, pi Hope this helped!
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