Solve `(3 sin theta-sin 3 theta)/(sin theta)+(cos 3 theta)/(cos theta)=1`.
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We have `(3 sin theta- sin 3 theta)/(sin theta)+(cos 3 theta)/(cos theta)=1` `:. (4 sin^(3) theta)/(sin theta)+((4 cos^(3) theta-3 cos theta))/(cos theta)=1` `rArr 4 sin^(2) theta+4 cos^(2) theta -3 =1` `rArr 1=1` This is always true except `theta= (npi)/2, n in Z` for which either `sin theta` or `cos theta` is zero.
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