Question

If the equation (1−λ)sin^2
x+(1+λ)sin x+(1−λ)=0 is having real roots, then solution of λ is ​

Answer 1

Step-by-step explanation:

We have,

$(1 - \lambda) \sin^{2} (x) + (1 + \lambda) \sin(x) + (1 - \lambda) = 0$

Since, it has real roots, so, its discriminant ≥ 0

Now,

$(1 + \lambda)^{2} - 4(1 - \lambda)^{2} \geqslant 0$

$\implies(1 + \lambda^{2} + 2\lambda) - 4 + 8\lambda - 4\lambda^{2} \geqslant 0$

$\implies \: - 3\lambda^{2} + 10\lambda - 3 \geqslant 0$

$\implies \: 3\lambda^{2} - 10\lambda + 3 \leqslant 0$

$\implies3\lambda^{2} - 9\lambda - \lambda + 3 \leqslant 0$

$\implies3\lambda(\lambda - 3) - 1(\lambda - 3) \leqslant 0$

$\implies(\lambda - 3)(3\lambda - 1) \leqslant 0$

$\implies \lambda \in [\frac{1}{3},3]$