Determine a general formula (or formulas) for the solutions to the equation. Then determine the specific solutions (if any) in the interval[0,2π).
2sin^2θ+3sinθ+1=0
another? Why not show how far you get with one of these?
Aside from the trig, it's just algebra I:
(2sinθ+1)(sinθ+1) = 0
...
To find the general formula for the solutions to the equation 2sin^2θ + 3sinθ + 1 = 0, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, the equation is 2sin^2θ + 3sinθ + 1 = 0. So, we have:
a = 2
b = 3
c = 1
Substituting these values into the quadratic formula, we get:
sinθ = (-3 ± √(3^2 - 4(2)(1))) / (2(2))
sinθ = (-3 ± √(9 - 8)) / 4
sinθ = (-3 ± √1) / 4
sinθ = (-3 ± 1) / 4
This gives us two possible solutions for sinθ:
1. sinθ = (-3 + 1) / 4 = -2/4 = -1/2
2. sinθ = (-3 - 1) / 4 = -4/4 = -1
To find the specific solutions in the interval [0, 2π), we need to find the corresponding values of θ.
1. sinθ = -1/2:
We know that sinθ = -1/2 corresponds to the reference angle π/6 in the second and third quadrants. So, we have:
θ = π - π/6 = 5π/6
θ = 2π + π/6 = 13π/6
2. sinθ = -1:
We know that sinθ = -1 corresponds to the reference angle π/2 in the third and fourth quadrants. So, we have:
θ = π - π/2 = π/2
θ = 2π + π/2 = 3π/2
Therefore, the specific solutions in the interval [0, 2π) for the equation 2sin^2θ + 3sinθ + 1 = 0 are:
θ = 5π/6, π/2, 3π/2, 13π/6