If the equation \sin^2 x−a\sin x+b=0 has only one solution

taoigas

taoigas

Answered question

2022-01-26

If the equation sin2xasinx+b=0 has only one solution in (0,π), then what is the range of b?

Answer & Explanation

Natalia Thomas

Natalia Thomas

Beginner2022-01-27Added 7 answers

If x(0,π) solves the equation then πx(0,π) also solves the equation since it holds sinx=sin(πx). By the uniqueness of solution on (0,π), we have x=πx, i.e. x=π2. Hence sin(π2)=1 solves the quadratic equation, which implies that
sin2(x)asinx+b=(sinx1)(sinxb)
In order that sinx=b has no solution on (0,π) other than x=π2, it must be that
b1    or    b0.

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