How do I solve for this equation? `2cos^2 theta-3cos theta +1=0,0 <= theta < 2pi`
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You should use the following substitution in the given equation, such that:
`cos theta = t => cos^2 theta = t^2`
`2t^2 - 3t + 1 = 0`
You need to split the midterm `-3t` into two terms such that:
`2t^2 - 2t - t + 1 = 0`
You need to group the terms such that:
`(2t^2 - 2t) - (t- 1) = 0`
Factoring out 2t in the first group yields:
`2t(t-1) - (t-1) = 0`
You need to factor out `(t-1)` such that:
`(t-1)(2t-1) = 0`
You need to solve the followings equations such that:
`t - 1 = 0 => t = 1 `
`2t - 1 = 0 => 2t = 1 => t = 1/2`
You need to solve for `theta in [0,2pi)` the following equations such that:
`cos theta = 1 => theta = 0`
`cos theta = 1/2 => theta = pi/3 and theta = 2pi-pi/3 = 5pi/3`
Hence, evaluating the solutions to the given equation, for `theta in [0,2pi), ` yields `theta = 0, theta = pi/3, theta = 5pi/3.`
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