Question

How do you find the general solutions for `cosx tanx - 2 cos^2x = -1`?

Answer 1

For `AAn in ZZ`
`color(white)("XXXX")x=pi/6+npi or (5pi)/6+npi or (3pi)/2+npi`

Explanation:

`cos(x)*tan(x) - 2cos^2(x) = -1`
`rarrcolor(white)("XXX")cancel(cos(x))*sin(x)/cancel(cos(x))-2 (1-sin^2(x)) = -1`

`rarrcolor(white)("XXX")2sin^2(x)+sin(x)-1 = 0`

`rarrcolor(white)("XXX")`(2sin(x)-1)*(sin(x)+1) = 0#

`rarrcolor(white)("XXX")sin(x) = 1/2 or sin(x)=-1`

Within `[0,2pi]`
`color(white)("XXX")sin(x)=1/2color(white)("XXX")rarrcolor(white)("XXX")x=pi/6 or x=(5pi)/6`

`color(white)("XXX")sin(x) = -1color(white)("XXX")rarrcolor(white)("XXX")x=(3pi)/2`