Question

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

Answer 1

`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]`)