Question

Find m to cos²x-(m²-3)sinx+2m²-3=0 have root

Answer 1

`-√(2) \le m \le √(2)` would ensure that at least one real root exists for this equation when solving for
`x`.

Explanation:

Apply the Pythagorean identity
`1 - \sin^(2)(x) = \cos^(2)(x)` to replace the cosine this equation with sine:

`(1 - \sin^(2)(x)) - (m^2 - 3)\, \sin(x) + 2\, m^2 - 3 = 0`.

Multiply both sides by
`(-1)` to obtain:

`-1 + \sin^(2)(x) + (m^2 - 3)\, \sin(x) - 2\, m^2 + 3 = 0`.

`\sin^(2)(x) + (m^2 - 3)\, \sin(x) - 2\, m^2 + 2 = 0`.

If
`y = \sin(x)`, then this equation would become a quadratic equation about
`y`:

`y^(2) + (m^2 - 3)\, y + (- 2\, m^2 + 2) = 0`.

• 
  `a = 1`.
• 
  `b = m^(2) - 3`.
• 
  `c = -2\, m^(2) + 2`.

However,
`-1 \le \sin(x) \le 1` for all real
`x`.

Hence, the value of
`y` must be between
`(-1)` and
`1` (inclusive) for the original equation to have a real root when solving for
`x`.

Determinant of this quadratic equation about
`y`:

`\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}`.

Hence, when solving for
`y`, the roots of
`y^(2) + (m^2 - 3)\, y + (- 2\, m^2 + 2) = 0` in terms of
`m` would be:

`\begin{aligned}y_1 &= \frac{-b + \sqrt{b^(2) - 4\, a\, c}}{2\, a} \\ &= \frac{-(m^(2) - 3) + \sqrt{(m^(2) + 1)^(2)}}{2} \\ &= (-(m^(2) - 3) + (m^(2) + 1))/(2) = 2\end{aligned}`.

`\begin{aligned}y_2 &= \frac{-b - \sqrt{b^(2) - 4\, a\, c}}{2\, a} \\ &= \frac{-(m^(2) - 3) - \sqrt{(m^(2) + 1)^(2)}}{2} \\ &= (-(m^(2) - 3) - (m^(2) + 1))/(2) \\ &= (-2\, m^(2) + 2)/(2) = -m^(2) + 1\end{aligned}`.

Since
`y = \sin(x)`, it is necessary that
`-1 \le y \le 1` for the original solution to have a real root when solved for
`x`.

The first solution,
`y_1`, does not meet the requirements. On the other hand, simplifying
`-1 \le y_2 \le 1`,
`-1 \le -m^(2) + 1 \le 1` gives:

`-2 \le -m^(2) \le 0`.

`0 \le m^(2) \le 2`.

`-√(2) \le m \le √(2)`.

In other words, solving
`y^(2) + (m^2 - 3)\, y + (- 2\, m^2 + 2) = 0` for
`y` would give a real root between
`-1 \le y \le 1` if and only if
`-√(2) \le m \le √(2)`.

On the other hand, given that
`y = \sin(x)` for the
`x` in the original equation, solving that equation for
`x\!` would give a real root if and only if
`-1 \le y \le 1`.

Therefore, the original equation with
`x` as the unknown has a real root if and only if
`-√(2) \le m \le √(2)`.