1
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Question

# Find the number of the solutions of the equation 1+cosx+cos2x+sinx+sin2x+sin3x=0 which satisfy the condition π2<∣∣∣3x−π2∣∣∣≤π.

A
1
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B
5
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C
4
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D
2
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Solution

## The correct option is A 1(1+cos(x)+cos(2x))+(sinx+sin2x+sin3x)=0(1+cos(x)+2cos2x−1)+(sinx+sin3x+sin2x)=0(2cos2(x)+cos(x))+(2sin2x.cosx+sin2x)+0cos(x)(2cos(x)+1)+sin2x(2cos(x)+1)=0(cos(x)+sin2x)(2cos(x)+1)=02cos(x)+1=0cos(x)=−12Hence x=2π3,4π3And cosx+sin2x=0cosx+2sin(x)cos(x)=0cos(x)[1+2sin(x)]=0cos(x)=0 and sin(x)=−12x=π2,3π2 and x=7π6,−π6.Now π2<|3x−π2|≤π|3x−π2|>π2Or 3x−π2>π2 and 3x−π2<−π2x<0 and x>π3 ...(i)and |3x−π2|≤π−π≤3x−π2≤π−2π≤6x−π≤π−π6≤x≤π3... (ii)Hence from i and ii−π6≤x<0Hence in the above range we get only one solution of the above equation, that is =−π6.Hence answer is 1.

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