`cos^2 (x) - 2cos(x) - 1 = 0, [0,pi]` Use a graphing utility to approximate the solutions (to three decimal places) of the equation in the given interval.

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Chapter 5, 5.3 - Problem 76 - Precalculus (3rd Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`cos^2(x)-2cos(x)-1=0`

using quadratic formula,

`cos(x)=(-(-2)+-sqrt((-2)^2-4*1*(-1)))/2`

`cos(x)=(2+-sqrt(8))/2`

`cos(x)=1+-sqrt(2)`

Solutions for cosx=1+`sqrt(2)` : None as cos(x) can not be greater than one for real solutions.

cos(x)=1-`sqrt(2)`  , x=arccos(1-`sqrt(2)` )

See the attached graph, Approximate x=2.000

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