How do you solve $2cos^2x=1$ in the interval [0,360]?

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Answer 1 · 2016-11-22T08:09:54

The solutions are $S={45,135,225,315}$

Let's rewrite the equation

$2cos^2x-1=0$

$(sqrt2cosx+1)(sqrt2cosx-1)=0$

Therefore,

$(sqrt2cosx+1=0$ , $=>$ , $cosx=-1/sqrt2$

$x=135$ and $x=225$

and $sqrt2cosx-1=0$ , $=>$ , $cosx=1/sqrt2$

$x=45$ and $x=315$

The solutions, $S in[0,360]$

So, $S={45,135,225,315}$

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