find the exact solutions (è in degrees)
sin^2è+2cosè=-2
change the sin^2 to 1-cos^2, then rearrage the equation into quadratic form, use the quadratic equaion.
To find the exact solutions (ê in degrees) for the equation sin^2ê + 2cosê = -2, you can follow these steps:
1. Use the identity sin^2ê = 1 - cos^2ê to rewrite the equation as:
1 - cos^2ê + 2cosê = -2
2. Rearrange the equation to get it in quadratic form:
-cos^2ê + 2cosê + 3 = 0
3. To solve this quadratic equation, you can use the quadratic formula given by:
ê = (-b ± √(b^2 - 4ac))/(2a)
In the given quadratic, a = -1, b = 2, and c = 3.
4. Substituting these values into the quadratic formula, you get:
ê = (-(2) ± √((2)^2 - 4(-1)(3)))/(2(-1))
Simplifying this, you have:
ê = (-2 ± √(4 + 12))/(-2)
ê = (-2 ± √16)/(-2)
ê = (-2 ± 4)/(-2)
5. Solve for ê by considering both the positive and negative square root:
ê = (-2 + 4)/(-2) = 2/-2 = -1
ê = (-2 - 4)/(-2) = -6/-2 = 3
Therefore, the exact solutions for ê in degrees are -1° and 3°.