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(sorry, first answer I posted is for question 6. what follows is the solution for question 7)
We want to solve for x in:
`2cos^2 x - cosx - 1 = 0`
We can solve this by transforming it to a quadratic formula with `u = cosx` .
Hence, we have `2u^2 - u - 1 = 0` . Using the quadratic formula:
`u = (1 pm sqrt(9))/4`
Hence, u is 1 or u is -1/2.
We know that`u = cosx` .
`cosx = 1`
`cosx = -1/2`
For the first case, `cos(x) = 1` means `x = cos^-1 (1) =0` . In fact, cosx = 1 when x is of the form `x = 2pin.`
Hence, the solutions we have from this are 0 and 360 degrees.
For the second case, `cos(x) = -1/2` means `x = cos^-1(-1/2).`
First, note that cosx here is negative. This means that the terminal side of the angle is in the second or third quadrant. If you solve this using a calculator, you will get 120 degrees. This is the angle in the second quadrant. The corresponding angle in the third quadrant is 240 degrees. We can get this by noting that 180 - 120 = 60. The equivalent of this angle on the negative y-axis will be 180 + 60 which is 240 degrees.
Now we only convert these into radians: `0 equiv 0, 120 equiv (2pi)/3, 240 equiv (4pi)/3, 360 equiv 2pi`
Hence, the answer is letter D.
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