Document2 - Arkansas Tech University

2sinx(2cosx v6(2cosx 0 (2 cos x — sm x 2cosx— — O cos x Il;r The solutions are 6 4 2sinx— — O 4 sm x 3m 11m 6 2sin2 x +1 2sm x—3smx+1 (2 sin x — x — 1) 3 smx 0 0 0 2' sin x — 1 sm x 6 2sinx —1 0 sm x 6 6 m The solutions are 6 . 4cos x cos 3 x 0 6 6 7m 11m cos x 3sinx — 5 3 sin x sm x ................
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