EddyBenzen122
New member
- Joined
- Jul 22, 2021
- Messages
- 28
The entire question: "Let f be the function f(θ) = 2cos2(θ) +4cosθ + 3, for -360° <= θ <= 360°.
a) Show that this function may be written as f(θ) = 4cos ²θ + 4cosθ + 1
b) Consider the equation f(θ) = 0, for -360° <=θ <= 360°
i) How many distinct values of cos θ which satisfy this equation.
ii) Find all values of θ which satisfy this equation.
c) Given that f(θ) = c is satisfied by only three values of θ,find the value of c.
I've done a),b), but not c)
For c) would it make sense if i do c = 4cos ²θ + 4cosθ + 1 because c = f(θ), and f(θ) = 4cos ²θ + 4cosθ + 1, if yes then how would i determine the three values to proceed to the next step?
a) Show that this function may be written as f(θ) = 4cos ²θ + 4cosθ + 1
b) Consider the equation f(θ) = 0, for -360° <=θ <= 360°
i) How many distinct values of cos θ which satisfy this equation.
ii) Find all values of θ which satisfy this equation.
c) Given that f(θ) = c is satisfied by only three values of θ,find the value of c.
I've done a),b), but not c)
For c) would it make sense if i do c = 4cos ²θ + 4cosθ + 1 because c = f(θ), and f(θ) = 4cos ²θ + 4cosθ + 1, if yes then how would i determine the three values to proceed to the next step?