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Solve : 5cos2theta+2cos^2theta/2+1=0,-pi...

Solve : `5cos2theta+2cos^2theta/2+1=0,-pi

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Changing all the values in terms of `cos theta`, we get
`5(2 cos^(2) theta-1)+(1+cos theta)+1=0`
or `10 cos^(2) theta+cos theta-3=0`
or `(5 cos theta+3) (2 cos theta-1) =0`
`rArr cos theta=1/2, (-3)/5`
`rArr theta=pi/3, - pi/3, cos^(-1) (-3/5)= pi- cos^(-1) 3/5`
and `-pi + cos^(-1) 3/5" "[ :' -pi lt theta lt pi]`
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