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Question

If cosx(1+sinx)=|tanx|, then find the number of solutions in [0,2π].


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Solution

Given:

cosx(1+sinx)=|tanx|

There will be two cases to check the solution of tanx in [0,2π]

Case 1: tanx0

cosx(1+sinx)=sinxcosx

cos2x=sin2x+sinx

1-sin2x=sin2x+sinx

2sin2x+sinx1=0

sinx=-1,12

sinx=-1 (not possible)

sinx=12

x=π6

Case 2: tanx<0

cosx(1+sinx)=-sinxcosx

cos2x=-sin2xsinx

1-sin2x=-sin2xsinx

sinx=-1 (Not possible)

Only one solution exists x=π6


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