Expert Answer to "What are possible values of k if sin..."

Elle Weber

Answered question

2022-05-14

What are possible values of k if

\sin^{2} x - k \sin x - 3 = 0

has exactly two distinct real roots in

[0, π]

Marco Meyer

Beginner2022-05-15Added 16 answers

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document Let f k ( x ) = sin 2 ⁡ x − k sin ⁡ x − 3
the map defined on

[0, π]

. As

f_{k} (π - x) = f_{k} (x)

has exactly two roots on

[0, π]

if and only if

f_{k}

has exactly one root on

I = [0, π / 2)

. On I, sinx is stricly increasing from 0 to 1. As

x \mapsto \sin x

is continuous,

f_{k}

will have exactly one root on

[0, π / 2)

is and only if

g_{k} (t) = t^{2} - k t - 3

has exactly one root on

[0, 1)

g_{k} (0) = - 3 < 0

, we must have

g_{k} (1) > 0

, i.e.

k < - 2

Conversely, suppose that

k < - 2

. We have

g_{k}^{'} (x) = 2 t - k > 0

. Which implies that

g_{k}

is strictly increasing from -3 to

- k - 2 > 0

and has only one root on

[0, 1)

Conclusion:

k < - 2

is the right answer.

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