We can write cos²θ as (1-sin²θ), so the equation becomes:


(1-sin²θ)-sinθ-1/4=0


We can simplify the equation further by multiplying both sides by 4 to get rid of the fraction:


4(1-sin²θ)-4sinθ-1=0


We can then distribute the 4 and simplify:


4-4sin²θ-4sinθ-1=0


-4sin²θ-4sinθ+3=0


We can solve for sinθ by using the quadratic formula:


sinθ = (-b ±√(b²-4ac))/2a


sinθ = (-(-4) ±√((-4)²-4(-4)(3)))/2(-4)


sinθ = (4 ±√16+48)/(-8)


sinθ = (4 ±√64)/(-8)


We can simplify further to get:


sinθ = (-1/2) or sinθ = (-3/4)


Now that we have sinθ, we can solve for cosθ by using the identity:


cos²θ + sin²θ = 1


cos²θ = 1 - sin²θ


Plugging in sinθ = (-1/2), we get:


cos²θ = 1 - (-1/2)²


cos²θ = 3/4


cosθ = ±√3/2


Plugging in sinθ = (-3/4), we get:


cos²θ = 1 - (-3/4)²


cos²θ = 7/16


cosθ = ±√7/4


Therefore, the solutions for θ are:


θ = π/6, 11π/6, 2π/3, 4π/3


The solutions to the equation cos²θ - sinθ - 1/4 = 0 are θ = π/6, 11π/6, 2π/3, 4π/3.