ehearts wrote:I HAVE A QUESTION!
TYPE OF QUESTION:maths,
YOUR QUESTION:
Solve the following equation for 0≤ θ ≤ 360
2(sin θ ) ^ 2 - 3 sin θ + 1 = 0
thank you !
This might be a bit late, but the idea is that you want to replace sin θ as x and solve it like a quadratic equation first. You can re-write the equation this way because sin θ is a function, so for any θ, you will get one and only one value of sin θ. Call that one value x.
If you re-write sin θ as x, you get the following:
2(x)^2 - 3x + 1 = 0
Now you can solve for x by factoring the equation:
(2x - 1)(x - 1) = 0
Now set each set of parentheses equal to zero and solve for x each time:
2x - 1 = 0 x - 1 = 0
2x = 1 x = 1
x = 1/2
(alternatively, you can use the quadratic equation to solve for x if factoring is not obvious to you)
Now that you have x = 1/2 and x = 1, replace sin θ with x and sole for θ:
sin θ = 1/2 sin θ = 1
θ = arcsin(1/2) θ = arcsin(1)
Now, just recall that arcsin(1/2) and arcsin(1) have fixed values with a domain of θ = [-90, 90].
If you use a calculator to solve this equation, you get the following:
arcsin(1/2) = 30 arcsin(1) = 90
However, you don't want values just between -90 and 90, you want values between 0 and 360. If you use the unit circle, you will notice that another angle will also give you arcsin(1/2), and that angle is 30 clockwise from the left horizontal, or 150 degrees (let me know if you need an explanation of why this is). There is no other angle that give you arcsin(1), also due to unit circle geometry.
Therefore, you have three angles between 0 and 360 that will solve this equation: 30, 90, and 150