[sin(142 + θ) - sin(θ)/cos^2(θ)] = 1.09 - how do I solve for θ?

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

if im not mistaking its

step-by-step explanation:  

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0