Solve 6 sin^(2)x - 5 cos x - 2 = 0 in the interval 0 ≤ x ≤ 2pi
since sin^2x = 1-cos^2x, that means
6(1-cos^2x) - 5cosx - 2 = 0
6 - 6cos^2x - 5cosx - 2 = 0
6cos^2x + 5cosx - 4 = 0
(3cosx+4)(2cosx-1) = 0
so, where do you have
cosx = -4/3 ? nowhere
where is cosx = 1/2 ?
To solve the equation 6sin^2(x) - 5cos(x) - 2 = 0, we can use the substitution sin^2(x) = 1 - cos^2(x), which allows us to rewrite the equation as a quadratic equation in terms of cos(x).
Let's substitute sin^2(x) = 1 - cos^2(x) into the equation:
6(1 - cos^2(x)) - 5cos(x) - 2 = 0
Simplify the equation:
6 - 6cos^2(x) - 5cos(x) - 2 = 0
-6cos^2(x) - 5cos(x) + 4 = 0
Now, this is a quadratic equation in terms of cos(x). Let's solve it.
To solve the quadratic equation, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a).
In this case, a = -6, b = -5, and c = 4.
Substituting the values into the quadratic formula:
cos(x) = (-(-5) ± √((-5)^2 - 4(-6)(4))) / (2(-6))
cos(x) = (5 ± √(25 + 96)) / (-12)
cos(x) = (5 ± √121) / (-12)
cos(x) = (5 ± 11) / (-12)
We have two possible solutions:
1. cos(x) = (5 + 11) / (-12) = 16 / (-12) = -4 / 3
2. cos(x) = (5 - 11) / (-12) = -6 / (-12) = 1 / 2
Now, let's find the corresponding values of x using the inverse cosine function (cos^-1):
1. For cos(x) = -4/3:
Since -1 ≤ cos(x) ≤ 1, there is no solution in the interval 0 ≤ x ≤ 2π.
2. For cos(x) = 1/2:
x = cos^-1(1/2)
Using the unit circle or a calculator, we find that:
x = π/3 and x = 5π/3.
Since the interval given is 0 ≤ x ≤ 2π, the solution to the equation 6sin^2(x) - 5cos(x) - 2 = 0 in this interval is:
x = π/3 and x = 5π/3.