A) The equation can be put in standard quadratic trigonometric equation form by letting cosx = y. The equation becomes 5y² + 4y - 1 = 0.
B) To factor the equation, we can use the quadratic formula:
y = (-B ± √(B² - 4AC)) / 2A
where A = 5, B = 4, and C = -1.
y = (-4 ± √(4² - 4(5)(-1))) / 2(5)
y = (-4 ± √(16 + 20)) / 10
y = (-4 ± √36) / 10
y = (-4 ± 6) / 10
Two possible values for y:
y₁ = (6-4) / 10 = 0.2
y₂ = (-6-4) / 10 = -1
C) Since cosx = y, we can find the solutions:
x = arccos(0.2) ≈ 1.37 radians or 78.69 degrees
x = arccos(-1) ≈ π radians or 180 degrees
So the solutions to the equation 5cos²x+4cosx=1 are x ≈ 1.37 radians and x ≈ π radians within the interval 0° ≤ x ≤ 2π.
Consider the equation 5cos²x+4cosx=1
A)put the equation in standard quadratic trigometric equation form.
B) use the quadratic formula to factor the equation.
C)What are the solutions to 2 decimal places, where 0°less then or equal to x less then or equal to 2pi?
1 answer
1 answer