How to solve 2cos^(2)x-sinx=1 ?

I do understand how I get sinx which is sinx=1/2 and sinx=-1 but I do not understand what I should do next. I know that I should look at the unit circle and when I do that I get 3pi/2, pi/6 and 5pi/6 but the answer is x=pi/6+n2pi/3

I do understand how I get sinx which is sinx=1/2 and sinx=-1 but I do not understand what I should do next. I know that I should look at the unit circle and when I do that I get 3pi/2, pi/6 and 5pi/6 but the answer is x=pi/6+n2pi/3

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#x=npi+(-1)^npi/6# or #x = 2npi-pi/2#

Explanation:

#cos^2x = 1- sin^2x#
#2cos^2x = 2 - 2sin^2x#

#:. 2cos^2x - sin x = 2 - 2 sin^2x - sin x = 1#

#2sin^2x + sin x - 1 = 0#

#2sin^2x + 2sinx - sinx -1 = 0#

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

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

# sin x = 1/2, -1#

If #sin x = 1/2=sin(pi/6)# and #x=npi+(-1)^npi/6# and in #[0,2pi)# #x=pi/6# or #(5pi)/6#

and if #x=-1=sin(-pi/2)# and #x = 2npi-pi/2# and in #[0,2pi)# #(3pi)/2#

Here #n# is an integer.

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