Solve sin 5 theta = cos 2 theta,0 < thet...

Solve `sin 5 theta = cos 2 theta`,0 < `theta` < `pi`.

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The correct Answer is:

`{((2n+(-1)^(n))pi//2)/(5+(-1)^(n)2):n in Z}`

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The solution of sin 2 theta + cos 2theta + sin theta + cos theta + 1 =0 in the first quadrant is

`2n pi+(2pi)/(3) : n in Z` or `(2n-1)pi + (pi)/(6) : n in Z`

`(2n +1)(pi)/(2) : n in Z` or `n pi - (pi)/(4) : n in Z`

`2n pi + (pi)/(2) : n in Z`

No solution

`2n pi : n in Z` or `(2n pi)/(3)-(pi)/(6) : n in Z`

`2n pi(pi)/(3) : n in Z` or `(2n pi)/(3) - (pi)/(6) : n in Z`

`2n pi : n in Z` or `(n pi)/(3) - (pi)/(6) : n in Z`

`2n pi(pi)/(3) : n in Z` or `(n pi)/(3) - (pi)/(6) : n in Z`

`{pi/3,(5pi)/3}`

`{pi,pi/2}`

`{pi/3,pi/4}`

None