Which function has real zeros at x = 3 and x = 7?
1 answer:
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The solutions to 1 - cos(x) = 2 - 2sin²(x) from (-π, π) are (-π/3, 0.5) and (π/3, 0.5)
<h3>How to solve the trigonometric equations?</h3><u>Equation 1: 1 - cos(x) = 2 - 2sin²(x) from (-π, π)</u>
The equation can be split as follows:
y = 1 - cos(x)
y = 2 - 2sin²(x)
Next, we plot the graph of the above equations (see graph 1)
Under the domain interval (-π, π), the curves of the equations intersect at:
(-π/3, 0.5) and (π/3, 0.5)
Hence, the solutions to 1 - cos(x) = 2 - 2sin²(x) from (-π, π) are (-π/3, 0.5) and (π/3, 0.5)
<u>Equation 2: 4cos⁴(x) - 5cos²(x) + 1 = 0 from [0, 2π)</u>
The equation can be split as follows:
y = 4cos⁴(x) - 5cos²(x) + 1
y = o
Next, we plot the graph of the above equations (see graph 2)
Under the domain interval [0, 2π), the curves of the equations intersect at:
(π/3, 0), (2π/3, 0), (π, 0), (4π/3, 0) and (5π/3, 0)
Hence, the solutions to 4cos⁴(x) - 5cos²(x) + 1 = 0 from [0, 2π) are (π/3, 0), (2π/3, 0), (π, 0), (4π/3, 0) and (5π/3, 0)
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