For the following equation, why cannot I not solve it? I keep getting an error.
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2cosx - 3 = 0
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I first isolated for "x"
2cosx-3 = 0
(2cosx/2) = (3/2)
x = cos^-1(3/2)
x = error? why?
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My textbook answer:
30 and 330 degrees
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Make sure you copied the question correctly.
The way it stands
cos x = 3/2 has indeed no solution
(the sine and cosine function does not exceed a value >1 or < -1
The reason you are getting an error when you try to solve the equation 2cosx - 3 = 0 and find x = cos^(-1)(3/2) is because the range of the inverse cosine function (cos^(-1)) is limited to values between -1 and 1. However, in this case, 3/2 is outside the range of valid values for the cosine function.
To solve this equation, you need to consider the range of the cosine function. The cosine function takes input angles (in radians or degrees) and outputs values between -1 and 1. Therefore, when trying to isolate cos(x) in this equation, you need to make sure that it falls within this range.
Starting from the equation:
2cos(x) - 3 = 0
First, add 3 to both sides of the equation:
2cos(x) = 3
Next, divide both sides by 2:
cos(x) = 3/2
Since 3/2 is outside the valid range for the cosine function, it means that the equation has no solutions. Therefore, you cannot find a real value for x that satisfies this equation.
The textbook answer you provided, which states that the solution is 30 and 330 degrees, seems to be incorrect. If you plug those values into the equation, you will see that they do not satisfy it (2cos(30) - 3 = -2 and 2cos(330) - 3 = -2).
In summary, the equation 2cosx - 3 = 0 has no real solutions.