Question - Solving Quadratic Equations in Terms of tan(x)
Solution:
The equation provided in the image is a quadratic equation in terms of tan(x):(tan(x))^2 - 1.3tan(x) - 0.48 = 0To solve it, we can use the quadratic formula:tan(x) = [-b ± sqrt(b^2 - 4ac)] / (2a)Here, a = 1, b = -1.3, and c = -0.48. Plugging these values into the quadratic formula gives us:tan(x) = [1.3 ± sqrt((1.3)^2 - 4(1)(-0.48))] / (2(1))tan(x) = [1.3 ± sqrt(1.69 + 1.92)] / 2tan(x) = [1.3 ± sqrt(3.61)] / 2tan(x) = [1.3 ± 1.9] / 2 This yields two possible solutions for tan(x):tan(x) = (1.3 + 1.9) / 2 = 3.2 / 2 = 1.6tan(x) = (1.3 - 1.9) / 2 = -0.6 / 2 = -0.3Now, we will find x for each of these tan(x) values using inverse trigonometric functions and considering the periodicity and symmetry of the tan function, which has a period of π.For tan(x) = 1.6, we can use the arctan function:x1 = arctan(1.6)Because tan(x) is positive in the first and third quadrants, and we're looking for solutions in the interval [0, 2π], we'll also consider:x2 = x1 + πThe solutions will be:x1 ≈ arctan(1.6)x2 ≈ arctan(1.6) + πFor tan(x) = -0.3, again we use the arctan function:x3 = arctan(-0.3)Since tan(x) is negative in the second and fourth quadrants and within [0, 2π], we will again add π to find an additional solution in the second quadrant:x4 = x3 + πThe solutions will be:x3 ≈ arctan(-0.3)x4 ≈ arctan(-0.3) + πYou will need to use a calculator to find the numeric values of x1, x3, and add π to find x2 and x4. Make sure that all solutions are within the given interval [0, 2π]. If any solution falls outside this interval, it must be adjusted using the periodicity of the tan function (by adding or subtracting integer multiples of π) to find the equivalent solution within the interval.
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