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Question

The general solution of sin2x2 cosx+14=0 is x=(nϵZ)

A
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Solution

The correct option is C
sin2x2cos x+14=04(1cos2 x)8cos x+1=04 cos2x+8 cosx5=04cos2 x+10 cos x2 cos x5=0(2 cos x+5)(2 cos x1)=0cos x=12=cosπ3x=2nπ±π3.

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