# Write down the number of roots of the equation 2cos 2θ-1 = 0 in the interval 0 ≤ θ ≤ 2π.

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### 1 Answer

`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`