Find all solutions on the interval (0, 2pi) of the equation: 2(sin^2)t-3sint+1=0
how do you get this one started?
let s=sin(t), so
2(sin^2)t-3sint+1=0
becomes
2s²-3s+1=0
(2s-1)(s-1)=0
=>
s=1/2 or s=1
=>
sin(x)=1/2 or sin(x)=1
Now solve for x for 0≤x≤2π
That makes so much sense. Thank you!
You're welcome!
To solve the equation 2(sin^2)t - 3sint + 1 = 0, you can use the quadratic formula since it is a quadratic equation in terms of sin(t). Here's how to get started:
1. Recognize that the equation can be simplified using the identity sin^2(t) = 1 - cos^2(t).
Substitute sin^2(t) with 1 - cos^2(t) in the equation:
2(1 - cos^2(t)) - 3sin(t) + 1 = 0
2. Rearrange the terms to form a quadratic equation in terms of cos(t):
2cos^2(t) - 3sin(t) + 1 - 2 = 0
2cos^2(t) - 3sin(t) - 1 = 0
3. Now, we can apply the quadratic formula:
The quadratic formula is given by x = (-b ± √(b^2 - 4ac)) / (2a),
where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.
In this case, a = 2, b = -3, and c = -1. So we have:
cos(t) = [3 ± √((-3)^2 - 4*2*(-1))] / (2*2)
cos(t) = [3 ± √(9 + 8)] / 4
cos(t) = [3 ± √17] / 4
4. Solve for cos(t) by taking the positive and negative square root of 17:
cos(t) = (3 + √17) / 4 or cos(t) = (3 - √17) / 4
5. Finally, find the solutions for t by taking the inverse cosine (arccos) of both sides:
t = arccos((3 + √17) / 4), which gives one solution.
and t = arccos((3 - √17) / 4), which gives another solution.
Note: Since the equation is defined on the interval (0, 2π), make sure to consider all possible solutions within this interval.