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using previous part


Deduce the number of roots of the equation 2cos 2θ-1 = 0 in the interval

10π ≤ θ ≤ 20π.

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`2cos 2theta-1 = 0 `

`rArr cos 2theta=1/2`

`rArr cos 2theta=cos(pi/3)`

`rArr 2theta=2kpi+-pi/3` where k can be any integer.

Hence, `theta=kpi+-pi/6`

For `k=10` , `theta=(61pi)/6` or `theta=(59pi)/6`

For` k=11` , `theta=(67pi)/6` or `theta=(65pi)/6`

For `k=12` , `theta=(73pi)/6` or `theta=(71pi)/6`

For `k=13` , `theta=(79pi)/6` or `theta=(77pi)/6`

For `k=14` , `theta=(85pi)/6` or `theta=(83pi)/6`

For `k=15` , `theta=(91pi)/6` or `theta=(89pi)/6`

For `k=16` , `theta=(97pi)/6` or `theta=(95pi)/6`

For `k=17` ,` theta=(103pi)/6` or `theta=(101pi)/6`

For` k=18 ` , `theta=(109pi)/6` or `theta=(107pi)/6`

For `k=19` , `theta=(115pi)/6` or `theta=(113pi)/6`

For `k=20` , `theta=(121pi)/6 ` or `theta=(119pi)/6`

Since the given interval is `10pi lt= theta lt= 20pi` , the roots of the equation are

`(61pi)/6,(65pi)/6,(67pi)/6,(71pi)/6,(73pi)/6, (77pi)/6,(79pi)/6,(83pi)/6,(85pi)/6,(89pi)/6,(91pi)/6,(95pi)/6, (97pi)/6,(101pi)/6,(103pi)/6,(107pi)/6,(109pi)/6,(113pi)/6, (115pi)/6,(119pi)/6.`

from the previous part :

For a range of 2pi, no of roots=4

Hence, For a range of 10pi, no of roots=20

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