Trigonometric EquationHard
Question
The equation (cos p -1)x2 + (cos p)x + sin p = 0 in the variable x, has real roots. Then p can take any value in the interval
Options
A.(0, 2π)
B.(- π, 0)
C.
D.(0, π)
Solution
Since, the giuve quadratic equation
(cos p - 1)x2 + (cos p)x + sin p = 0
has real roots.
∴ Discriminant, cos2 p - 4sin p(cos p -1) ≥ 0
⇒ (cos p - 2sin p)2 - 4sin2 p + 4sin p ≥ 0
⇒ (cos p - 2sin p)2 + 4sin p(- sin p) ≥ 0
∵ 4sin p(1 - sin p) > 0 for 0 < p < π
and (cos p - 2sin p)2 ≥ 0
Thus, (cos p - 2sin p)2 + 4sin p(1 - sin p) ≥ 0
for 0 < p < π
Hebce, the equation has real roots for 0 < p < π.
(cos p - 1)x2 + (cos p)x + sin p = 0
has real roots.
∴ Discriminant, cos2 p - 4sin p(cos p -1) ≥ 0
⇒ (cos p - 2sin p)2 - 4sin2 p + 4sin p ≥ 0
⇒ (cos p - 2sin p)2 + 4sin p(- sin p) ≥ 0
∵ 4sin p(1 - sin p) > 0 for 0 < p < π
and (cos p - 2sin p)2 ≥ 0
Thus, (cos p - 2sin p)2 + 4sin p(1 - sin p) ≥ 0
for 0 < p < π
Hebce, the equation has real roots for 0 < p < π.
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