consider the equation cos^2x + 4cosx =1
a) put the equation in standard quadratic trigonometric equation form.
b) use the quadratic formula to factor the equation.
c) what are the solutions to two decimal places, where 0 degrees < x < 2pi
a) To put the equation in standard quadratic trigonometric equation form, we can rearrange it:
cos^2x + 4cosx = 1
cos^2x + 4cosx - 1 = 0
b) Now we can use the quadratic formula to factor the equation:
The quadratic formula states that for a quadratic equation of the form ax^2 + bx + c = 0, the solutions can be found using the formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
In our case, a = 1, b = 4, and c = -1. Plugging these values into the quadratic formula, we get:
x = (-4 ± sqrt(4^2 - 4(1)(-1))) / 2(1)
Simplifying:
x = (-4 ± sqrt(16 + 4)) / 2
x = (-4 ± sqrt(20)) / 2
x = (-4 ± 2sqrt(5)) / 2
x = -2 ± sqrt(5)
So the solutions are x = -2 + sqrt(5) and x = -2 - sqrt(5).
c) To find the solutions to two decimal places, we can evaluate the two values for x:
x = -2 + sqrt(5) ≈ 0.24
x = -2 - sqrt(5) ≈ -4.24
Therefore, the solutions to two decimal places are x ≈ 0.24 and x ≈ -4.24, where 0 degrees < x < 2pi.