The solutions of sin^2 x + cos x =1 with 0 less than or equal to x less than or equal to 2pi are?

Question

im trying to study for a test but i really dont get how to do this? please help!!!

Answer 1

Recall the trig identity:

sin^2(x)+cos^2(x)=1

So sin^2(x)=1-cos^2(x)

We substitute:

1-cos^2(x)+cos(x)=1

cos^2(x)-cos(x)=0

cos(x) (cos(x)-1)=0

cos(x)=0

x=pi/2, 3pi/2

cos(x)-1=0

x=0, 2pi

Answers: x=0, pi/2, 3pi/2, 2pi

Answer 2

sin^2 x + cos x = 1
1 - cos^2 x + cos x = 1
- cos^2 x + cos x = 0
cos^2 x - cos x = 0
cos x(cos x - 1) = 0
cos x = 0
x = π/2, 3π/2
cos x = 1
x = 0, 2π

x = 0, π/2, 3π/2, 2π

Answer 3

1 - cos ² x + cos x = 1
cos ² x - cos x = 0
cos x (cos x - 1) = 0
cos x = 0 , cos x = 1
x = π/2 , 3π/2 , 0 , 2π

Answer 4

Woops

Answer 5

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