Solve each equation for 0<_ 0<_ 2pi(3.14)

A) 2cos^20 - sin0 -1=0
My answers: sin0= -1/2 or sin0=1 so 0 = 7pi/6, 11pi/6 and pi/2

But the back of the book says: pi/6, 5pi/6 and 3pi/2

2-2sin^2-sin-1=0

2sin^2+sin-1 = 0
(2sin-1)(sin+1) = 0
sin = 1/2 or -1
sin = 1/2 at pi/6 and 5pi/6
sin = -1 at 3pi/2

Better check your factoring.
Your angles are correct for the values you obtained, but those values had the wrong signs, and hence, the wrong sines :-)

To solve the equation 2cos^2(θ) - sin(θ) - 1 = 0, you need to manipulate the equation to isolate the trigonometric function and then use trigonometric identities to solve for θ.

Step 1: Rearrange the equation so that one side equals zero:
2cos^2(θ) - sin(θ) - 1 = 0

Step 2: Combine like terms:
2cos^2(θ) - sin(θ) = 1

Step 3: Identify a suitable trigonometric identity to manipulate the equation. In this case, the Pythagorean identity is helpful:
sin^2(θ) + cos^2(θ) = 1

Step 4: Substitute the identity into the equation:
2(1 - sin^2(θ)) - sin(θ) = 1
2 - 2sin^2(θ) - sin(θ) = 1
-2sin^2(θ) - sin(θ) + 1 = 0

Step 5: Rearrange the equation to match the quadratic form:
-2sin^2(θ) - sin(θ) + 1 = 0
-2sin^2(θ) - sin(θ) = -1

Step 6: Multiply both sides of the equation by -1 to switch the signs:
2sin^2(θ) + sin(θ) = 1

Step 7: Factor the equation:
sin(θ)(2sin(θ) + 1) = 1

Step 8: Set each factor equal to zero and solve for θ:
sin(θ) = 0 or 2sin(θ) + 1 = 0

For sin(θ) = 0, the angle θ can be 0, π, or any other multiple of π. These values fall within the given range of 0 ≤ θ ≤ 2π.

For 2sin(θ) + 1 = 0, we subtract 1 from both sides and then divide by 2 to isolate the sine function:
2sin(θ) = -1
sin(θ) = -1/2

To find the angles corresponding to sin(θ) = -1/2, you can refer to the unit circle or the trigonometric ratios for specific angles. In this case, the angles in the range 0 ≤ θ ≤ 2π that satisfy sin(θ) = -1/2 are π/6, 5π/6, and 3π/2.

Therefore, the correct solutions to the equation are θ = 0, π, π/6, 5π/6, and 3π/2.