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Question

The equation (cosp1)x2+(cosp)x+sinp=0 has real roots, then 'P' can take any value in the interval

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Solution

(cosp1)x2+(cosp)x+(sinp)=0
As roots are real, discriminant0
0b24ac0
cos2p4(cosp1)sinp0
cos2p4cospsinp+4sinp0
cos2p4cospsinp+4sin2p4sin2p+4sinp by adding and subtracting 4sin2p
(cosp2sinp)2+4sinp(1sinp)0
cosp2sinp is always positive.
As 1sinp1
1sinp1
1+11sinp0 by adding 1
21sinp0
As sinp0 sinp is positive and is in first and third quadrant.
p(0,π) in first and second quadrant in anticlockwise direction.

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