Question

$\sin^4x + \cos^4x + \sin2x + \alpha = 0$

Range of $\alpha = ?$

Answer 1

$\left[-\frac{3}{2}, \frac{1}{2}\right]$

Answer 2

To find the range of $\alpha$ for which the equation $\sin^4x + \cos^4x + \sin2x + \alpha = 0$ has real solutions for $x$, we first simplify the trigonometric expression.

Step 1: Simplify $\sin^4x + \cos^4x$

We know the identity $a^2 + b^2 = (a+b)^2 - 2ab$. Let $a = \sin^2x$ and $b = \cos^2x$.

$\sin^4x + \cos^4x = (\sin^2x + \cos^2x)^2 - 2\sin^2x\cos^2x$

Since $\sin^2x + \cos^2x = 1$, this simplifies to:

$\sin^4x + \cos^4x = 1^2 - 2(\sin x\cos x)^2$

We also know that $\sin2x = 2\sin x\cos x$, which means $\sin x\cos x = \frac{\sin2x}{2}$.

Substitute this into the expression:

$\sin^4x + \cos^4x = 1 - 2\left(\frac{\sin2x}{2}\right)^2$

$\sin^4x + \cos^4x = 1 - 2\frac{\sin^22x}{4}$

$\sin^4x + \cos^4x = 1 - \frac{\sin^22x}{2}$

Step 2: Substitute the simplified expression back into the original equation

The given equation becomes:

$\left(1 - \frac{\sin^22x}{2}\right) + \sin2x + \alpha = 0$

Step 3: Introduce a substitution

Let $y = \sin2x$. Since $x$ is a real number, $\sin2x$ can take any value in its standard range, i.e., $y \in [-1, 1]$.

The equation transforms into an algebraic equation in terms of $y$:

$1 - \frac{y^2}{2} + y + \alpha = 0$

Step 4: Express $\alpha$ in terms of $y$

Rearrange the equation to isolate $\alpha$:

$\alpha = \frac{y^2}{2} - y - 1$

Step 5: Find the range of $\alpha$ by analyzing the function $f(y) = \frac{y^2}{2} - y - 1$ for $y \in [-1, 1]$

This is a quadratic function $f(y) = ay^2 + by + c$ with $a = 1/2$, $b = -1$, and $c = -1$. Since the coefficient of $y^2$ ($a = 1/2$) is positive, the parabola opens upwards. The vertex of the parabola is located at $y = -\frac{b}{2a}$.

$y_{vertex} = -\frac{-1}{2(1/2)} = -\frac{-1}{1} = 1$.

The interval for $y$ is $[-1, 1]$. The vertex $y=1$ is included in this interval and is at its right boundary. For an upward-opening parabola, the minimum value within an interval occurs either at the vertex or at one of the endpoints. Since the vertex is at $y=1$, this is where the minimum value occurs within the interval $[-1, 1]$.

Minimum value of $f(y)$ at $y=1$:

$f(1) = \frac{(1)^2}{2} - (1) - 1 = \frac{1}{2} - 1 - 1 = \frac{1}{2} - 2 = -\frac{3}{2}$.

The maximum value of $f(y)$ within the interval $[-1, 1]$ will occur at the endpoint furthest from the vertex. The vertex is at $y=1$. The distance from $y=1$ to $y=-1$ is $|1 - (-1)| = 2$. The distance from $y=1$ to $y=1$ is $0$. Thus, the maximum value occurs at $y=-1$.

Maximum value of $f(y)$ at $y=-1$:

$f(-1) = \frac{(-1)^2}{2} - (-1) - 1 = \frac{1}{2} + 1 - 1 = \frac{1}{2}$.

Therefore, the range of the function $f(y) = \frac{y^2}{2} - y - 1$ for $y \in [-1, 1]$ is $\left[-\frac{3}{2}, \frac{1}{2}\right]$. This is the required range for $\alpha$.