What are all the solutions between 0 and 2π for 2cos^2 = sinx+1?

1 Answer
Alan P.
Mar 29, 2016

x in {pi/6, (5pi)/6, (3pi)/2}

Explanation:

Assuming the equation was meant to be:
color(white)("XXX")2cos^2(color(red)(x))=sin(x)+1

Using the relationship:
color(white)("XXX")cos^2(x)+sin^2(x)=1
2 cos^2(x)=sin(x)+1

rarr 2(1-sin^2(x))=sin(x)+1

rarr 2sin^2(x)+sin(x)-1=0

rarr (2sin(x)-1)(sin(x)+1)=0

rarr sin(x)=1/2color(white)("XXX")orcolor(white)("XXX")sin(x)=-1
rarr x=pi/6 or (5pi)/6color(white)("XXXXXXX")rarr x=(3pi)/2
color(white)("XXXXXXXXXX")(for x in[0,2pi])

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