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Sagot :
Answer:
[tex]-\sqrt{2} \le m \le \sqrt{2}[/tex] would ensure that at least one real root exists for this equation when solving for [tex]x[/tex].
Step-by-step explanation:
Apply the Pythagorean identity [tex]1 - \sin^{2}(x) = \cos^{2}(x)[/tex] to replace the cosine this equation with sine:
[tex](1 - \sin^{2}(x)) - (m^2 - 3)\, \sin(x) + 2\, m^2 - 3 = 0[/tex].
Multiply both sides by [tex](-1)[/tex] to obtain:
[tex]-1 + \sin^{2}(x) + (m^2 - 3)\, \sin(x) - 2\, m^2 + 3 = 0[/tex].
[tex]\sin^{2}(x) + (m^2 - 3)\, \sin(x) - 2\, m^2 + 2 = 0[/tex].
If [tex]y = \sin(x)[/tex], then this equation would become a quadratic equation about [tex]y[/tex]:
[tex]y^{2} + (m^2 - 3)\, y + (- 2\, m^2 + 2) = 0[/tex].
- [tex]a = 1[/tex].
- [tex]b = m^{2} - 3[/tex].
- [tex]c = -2\, m^{2} + 2[/tex].
However, [tex]-1 \le \sin(x) \le 1[/tex] for all real [tex]x[/tex].
Hence, the value of [tex]y[/tex] must be between [tex](-1)[/tex] and [tex]1[/tex] (inclusive) for the original equation to have a real root when solving for [tex]x[/tex].
Determinant of this quadratic equation about [tex]y[/tex]:
[tex]\begin{aligned} & b^{2} - 4\, a\, c \\ =\; & (m^{2} - 3)^{2} - 4 \cdot (-2\, m^{2} + 2) \\ =\; & m^{4} - 6\, m^{2} + 9 - (-8\, m^{2} + 8) \\ =\; & m^{4} - 6\, m^{2} + 9 + 8\, m^{2} - 8 \\ =\; & m^{4} + 2\, m^{2} + 1 \\ =\; &(m^2 + 1)^{2} \end{aligned}[/tex].
Hence, when solving for [tex]y[/tex], the roots of [tex]y^{2} + (m^2 - 3)\, y + (- 2\, m^2 + 2) = 0[/tex] in terms of [tex]m[/tex] would be:
[tex]\begin{aligned}y_1 &= \frac{-b + \sqrt{b^{2} - 4\, a\, c}}{2\, a} \\ &= \frac{-(m^{2} - 3) + \sqrt{(m^{2} + 1)^{2}}}{2} \\ &= \frac{-(m^{2} - 3) + (m^{2} + 1)}{2} = 2\end{aligned}[/tex].
[tex]\begin{aligned}y_2 &= \frac{-b - \sqrt{b^{2} - 4\, a\, c}}{2\, a} \\ &= \frac{-(m^{2} - 3) - \sqrt{(m^{2} + 1)^{2}}}{2} \\ &= \frac{-(m^{2} - 3) - (m^{2} + 1)}{2} \\ &= \frac{-2\, m^{2} + 2}{2} = -m^{2} + 1\end{aligned}[/tex].
Since [tex]y = \sin(x)[/tex], it is necessary that [tex]-1 \le y \le 1[/tex] for the original solution to have a real root when solved for [tex]x[/tex].
The first solution, [tex]y_1[/tex], does not meet the requirements. On the other hand, simplifying [tex]-1 \le y_2 \le 1[/tex], [tex]-1 \le -m^{2} + 1 \le 1[/tex] gives:
[tex]-2 \le -m^{2} \le 0[/tex].
[tex]0 \le m^{2} \le 2[/tex].
[tex]-\sqrt{2} \le m \le \sqrt{2}[/tex].
In other words, solving [tex]y^{2} + (m^2 - 3)\, y + (- 2\, m^2 + 2) = 0[/tex] for [tex]y[/tex] would give a real root between [tex]-1 \le y \le 1[/tex] if and only if [tex]-\sqrt{2} \le m \le \sqrt{2}[/tex].
On the other hand, given that [tex]y = \sin(x)[/tex] for the [tex]x[/tex] in the original equation, solving that equation for [tex]x\![/tex] would give a real root if and only if [tex]-1 \le y \le 1[/tex].
Therefore, the original equation with [tex]x[/tex] as the unknown has a real root if and only if [tex]-\sqrt{2} \le m \le \sqrt{2}[/tex].
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