The values in the interval [0, 2π) which the functions f(x) = 2cos2θ and g(x) = − 1 + 4cos θ − 2cos2θ intersect are θ = 24.31° and θ = 65.71°
The functions f(x) = 2cos2θ and g(x) = −1 + 4cos θ − 2cos2θ intersect when
f(x) = g(x)
So, 2cos2θ = − 1 + 4cosθ − 2cos2θ
2cos2θ + 2cos2θ = − 1 + 4cosθ
4cos2θ = -1 + 4cosθ
Since cos2θ = 2cos²θ - 1
So, substituting this into the equation, we have
4cos2θ = -1 + 4cosθ
4(2cos²θ - 1) = - 1 + 4cosθ
8cos²θ - 4 = - 1 + 4cosθ
8cos²θ - 4cosθ - 4 + 1 = 0
8cos²θ - 4cosθ - 3 = 0
Let cosθ = y
So, 8y² - 4y - 3 = 0
For a quadratic equation ax + bx + c = 0, using the quadratic formula, x =[tex]x = \frac{-b +/- \sqrt{b^{2} - 4ac} }{2a}[/tex]
So from the above equation, a = 8, b = - 4 and c = - 3.
So,
[tex]y = \frac{-(-4) +/- \sqrt{4^{2} - 4X8X(-3)} }{2X8}\\y = \frac{-(-4) +/- \sqrt{16 + 96} }{16}\\y = \frac{4 +/- \sqrt{112} }{16}\\y = \frac{4 +/- 10.58 }{16}\\y = \frac{4 + 10.58 }{16} or \frac{4 - 10.58 }{16}\\y = \frac{14.58 }{16} or \frac{-6.58 }{16}\\y = 0.9113 or 0.4113[/tex]
Since cosθ = y
cosθ = 0.9113 or cosθ = 0.4113
θ = cos⁻¹(0.9113) or θ = cos⁻¹(0.4113)
θ = 24.31° or θ = 65.71°
So, the values in the interval [0, 2π) which the functions f(x) = 2cos2θ and g(x) = − 1 + 4cos θ − 2cos2θ intersect are θ = 24.31° and θ = 65.71°
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