Question #1ab02

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Answer 1 · 2018-01-30T17:17:22

# theta in {mpi/4 : m in ZZ}uu{(2n+1)pi/10 : n in ZZ}#.

Recall that, #sinx=siny rArr x=(-1)^ky+kpi, k in ZZ#.

#:. sin9theta=sintheta rArr 9theta=(-1)^ktheta+kpi, kin ZZ#.

If # k in ZZ" is even, say "k=2m#, then,

#9theta=(-1)^(2m)theta+2mpi=theta+2mpi, or, #

#8theta=2mpi rArr theta=mpi/4, m in ZZ#.

In case #k in ZZ" is odd, say "k=2n+1#, then,

#9theta=(-1)^(2n+1)theta+(2n+1)pi=-theta+(2n+1)pi, or, #

#10theta=(2n+1)pi rArr theta=(2n+1)pi/10, n in ZZ#.#

Altogether, the Solution Set is given by,

#{mpi/4 : m in ZZ}uu{(2n+1)pi/10 : n in ZZ}#.