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Question

Number of roots of the equation (3+cosx)2=42sin8x,xϵ[0,5π]are

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Solution

(3+cosx)2=42sin8xwherex[0,5π]

(3+cosx)2[4,16]as1sinx1,1cosx1

42sin8x[2,4]

Sobothcanbeequalonlywhenbothareequalto4

Thensinx=0andcosx=1atthesametime.

x=(0,2π,4π}

Wehavethreesolutions

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