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given,
sin2A=cos3A
2 sinA cosA=4cos3A-3cosA
2 sin A=4 cos2A-3
2 sin A =4(1-sin2A)-3
now solve for sinA
sin A=(-1-√5)/4 or (-1+√5)/4
Put A=18
and for 0
sin18=(-1+√5)/4
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