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Write down the number of roots of the equation 2cos 2θ-1 = 0 in the interval 0 ≤ θ...
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`2cos2theta-1 = 0`
`cos2theta = 1/2`
`cos2theta = cos(pi/3)`
`2theta = 2npi+-pi/3 ` where `n in Z`
`theta = npi+-pi/6`
When n = -1 then `theta = -(5pi)/6` OR `theta = -(7pi)/6`
When n = 0 then `theta = -pi/6` OR `theta = pi/6`
When n = 1 then `theta = (7pi)/6` OR `theta = (5pi)/6`
When n = 2 then `theta = (13pi)/6` OR `theta = (11pi)/6`
We need the answers in the range of `0-2pi` .
The answers are
`theta = pi/6, (7pi)/6, (5pi)/6,(11pi)/6`
Posted by jeew-m on August 31, 2013 at 11:34 AM (Answer #1)
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