Find the solutions to sin2(x) + cos(x) = 1, keeping 0 ≤ x < 2π

Sin2(x) +cos(x)=1
from the relation: (sin2(x) +cos2(x) =1 )
so , sin2(x)=1-cos2(x)
by subs. in the main eqn.
1-cos2(x) + cos(x) =1
by simplify the eqn.
cos(x) -cos2(x)=0
take cos(x) as a common factor
cos(x)* (1-cos(x))=0
then cos(x)=0 && cos(x)=1
cos(x)=0 if x= pi/2
& cos(x) = 1 if x = 0 , 2*pi
so the solution is x= {0,pi/2 , 2*pi}

Which statements did you include in your answer?

Isolate sin(x) by adding 4 and taking the square root of both sides.

State that sin(x) = 2 or sin(x) = –2.

State that –2 and 2 are undefined values  of the inverse sine function.

There are no solutions because –2 and 2 are not in the domain of the function.

Step-by-step explanation:

Hello here is a solution 
look graph