Q.One value of $\theta$ which satisfies the equation $\sin^4\theta - 2\sin^2\theta - 1$ lies between $0$ and $2\pi$.

The key idea is to treat the given equation as a quadratic in $\sin^2\theta$ and then verify if the solutions for $\sin^2\theta$ fall within its permissible range.

1. Let $x = \sin^2\theta$. The equation becomes a quadratic: $x^2 - 2x - 1 = 0$.
2. Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we solve for $x$: $x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$.
3. Thus, $\sin^2\theta = 1 + \sqrt{2}$ or $\sin^2\theta = 1 - \sqrt{2}$.
4. We know that for any real $\theta$, $0 \le \sin^2\theta \le 1$. …

The equation $\sin^4\theta - 2\sin^2\theta - 1 = 0$ can be simplified by recognizing its structure. Notice that all terms involve $\sin\theta$ raised to an even power. This suggests that we can treat $\sin^2\theta$ as a single variable, transforming the equation into a more familiar quadratic form.

The core idea is to make a substitution, solve the resulting algebraic equation, and then critically evaluate the solutions in the context of the original trigonometric function. For any real angle $\theta$, the value of $\sin\theta$ must be between $-1$ and $1$. This fundamental property implies that $\sin^2\theta$ must always lie in the interval $[0, 1]$. Any solution for $\sin^2\theta$ that falls outside this range is not physically possible for a real angle $\theta$.

Here is the step-by-step solution:

1. Substitute to form a quadratic equation. Let $x = \sin^2\theta$. It is important to remember the range constraint for $x$: since $-1 \le \sin\theta \le 1$, squaring this inequality gives $0 \le \sin^2\theta \le 1$. Therefore, any valid solution for $x$ must satisfy $0 \le x \le 1$. Substituting $x$ into the given equation $\sin^4\theta - 2\sin^2\theta - 1 = 0$, we get:

2. Solve the quadratic equation for $x$. We use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, for an equation of the form $ax^2 + bx + c = 0$. In our case, $a=1$, $b=-2$, and $c=-1$.

This gives us two potential values for $x$:
$x_1 = 1 + \sqrt{2}$
$x_2 = 1 - \sqrt{2}$

3. Check the validity of the solutions for $x$.

As established in Step 1, for $x = \sin^2\theta$ to correspond to a real angle $\theta$, $x$ must be in the range $[0, 1]$.

*   Consider $x_1 = 1 + \sqrt{2}$:
    Since $\sqrt{2} \approx 1.414$, $x_1 \approx 1 + 1.414 = 2.414$.
    This value is greater than $1$, which means it is outside the valid range for $\sin^2\theta$. Thus, $x_1 = 1 + \sqrt{2}$ is not a valid solution.

*   Consider $x_2 = 1 - \sqrt{2}$: …