The set of all values of λ for which the equation cos² 2x - 2sin⁴ x - 2cos² x = λ has a solution is

Q: Wait, why do we substitute $t = \cos^2 x$? Why not just solve for $x$ directly?

Solving for $x$ directly in this equation would lead to a nightmare of transcendental functions. By substituting $t = \cos^2 x$, we transform a complex trigonometric equation into a simple quadratic function. This is the 'Art of Reduction'—we map the problem into a domain where we have powerful tools like vertex analysis and range calculation.

Q: Why is the domain of $t$ restricted to $[0, 1]$?

Great catch! Since $t = \cos^2 x$, and we know that for any real $x$, the value of $\cos x$ oscillates between $-1$ and $1$, squaring it forces the value to stay between $0$ and $1$. If you ignore this, you might calculate the range of the parabola over its entire domain $(-\infty, \infty)$, which would give you the wrong answer.

Q: How do I know if the maximum occurs at the vertex or the boundaries?

Always look at the vertex of your parabola. If the vertex $t = 1/2$ lies within your interval $[0, 1]$, the minimum (for an upward-opening parabola) is at the vertex. The maximum, however, will always be at the boundary point furthest from the vertex. Here, both $t=0$ and $t=1$ are equidistant from $1/2$, so they both yield the same maximum value.

Q: Is it possible for $\lambda$ to be outside the range $[-3/2, -1]$?

If $\lambda$ were outside this range, the equation would have no real solution for $x$. The range we found represents all possible values the expression can take. If you pick a $\lambda$ outside this, you are essentially asking the parabola to intersect a horizontal line that is either below its lowest point or above its highest point—which is geometrically impossible!

Q: Do I need to worry about the $\sin^4 x$ term being negative?

Not at all! Because the power is even, $\sin^4 x$ is always non-negative. When you convert it to $(1 - \cos^2 x)^2$, you are essentially working with $(1-t)^2$. The algebra handles the signs for you automatically. Trust the substitution!

Question

The set of all values of $\lambda$ for which the equation $\cos^2 2x - 2\sin^4 x - 2\cos^2 x = \lambda$ has a solution is

Options:

(A) $[-2, -1]$

(B) $[-2, -\frac{3}{2}]$

(C) $[-1, -\frac{1}{2}]$

(D) $[-\frac{3}{2}, -1]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we aren't just solving an equation; we are peeling back the layers of a trigonometric onion.

At first glance, the expression $\cos^2 2x - 2\sin^4 x - 2\cos^2 x = \lambda$ looks like a chaotic mess of powers and double angles. But remember, in mathematics, chaos is often just order waiting to be discovered.

### The Unification

Imagine you are standing in a forest where every tree is a different species—some are $\cos 2x$, some are $\sin^4 x$, and some are $\cos^2 x$. To understand the forest, we need a common language.

Our goal is to unify these terms. We know the identity $\cos 2x = 2\cos^2 x - 1$. This is our bridge!

By substituting this, and using the fundamental identity $\sin^2 x = 1 - \cos^2 x$, we can rewrite everything in terms of a single variable: $t = \cos^2 x$.

### The Transformation

Now, let's perform the substitution. We define $t = \cos^2 x$. Because $x$ is a real number, $\cos x$ is trapped between $-1$ and $1$.

Therefore, $t$ is strictly confined to the interval $t \in [0, 1]$. This is a crucial realization—if you forget this boundary, you lose the soul of the problem.

Substituting $t$ into our equation, we get:

$$\lambda = (2t - 1)^2 - 2(1 - t)^2 - 2t$$

Take a deep breath. This looks like a standard algebraic expansion. Let's expand the terms carefully:

$$(2t - 1)^2 = 4t^2 - 4t + 1$$

$$-2(1 - t)^2 = -2(1 - 2t + t^2) = -2 + 4t - 2t^2$$

When we combine these with the remaining $-2t$, the magic happens. The linear terms $-4t$ and $+4t$ cancel out, leaving us with a beautiful, clean quadratic:

$$\lambda = 2t^2 - 2t - 1$$

### The Geometry of the Parabola

We have reduced a complex trigonometric problem to a simple parabola: $f(t) = 2t^2 - 2t - 1$. To find the range of $\lambda$, we need to find the range of this function over the interval $t \in [0, 1]$.

Let's complete the square to find the vertex:

$$f(t) = 2(t^2 - t) - 1 = 2\left(t^2 - t + \frac{1}{4} - \frac{1}{4}\right) - 1 = 2\left(t - \frac{1}{2}\right)^2 - \frac{3}{2}$$

This is a parabola opening upwards with its vertex at $t = 1/2$. Since $1/2$ is inside our interval $[0, 1]$, the minimum value of the function must occur at this vertex.

Plugging $t = 1/2$ into our equation, we find the minimum value of $\lambda$ is $-3/2$.

### The Boundaries and Victory

Finally, we check the boundaries of our interval, $t=0$ and $t=1$.

For $t=0$: 
$$f(0) = 2\left(0 - \frac{1}{2}\right)^2 - \frac{3}{2} = 2\left(\frac{1}{4}\right) - \frac{3}{2} = \frac{1}{2} - \frac{3}{2} = -1$$

For $t=1$: 
$$f(1) = 2\left(1 - \frac{1}{2}\right)^2 - \frac{3}{2} = 2\left(\frac{1}{4}\right) - \frac{3}{2} = \frac{1}{2} - \frac{3}{2} = -1$$

Both boundaries yield $-1$. Thus, as $t$ travels from $0$ to $1$, the value of $\lambda$ travels from $-3/2$ up to $-1$.

The range of $\lambda$ for which the equation holds true is **$[-\frac{3}{2}, -1]$**.

Animated Solution for Mathematics - Trigonometry: The set of all values of $λ$ for which the equation $cos^{2} 2 x - 2 sin^{4} x - 2 cos^{2} x = λ$ has a solution is

Select Answer:

Visualized Solution

Analyze the Equation

Apply Trigonometric Identities

Substitute $t = cos^{2} x$

Expand $(2 t - 1)^{2}$

Expand $- 2 (1 - t)^{2}$

Simplify the Expression

Complete the Square

Visualize the Function

Find Minimum Value

Find Maximum Value

Final Range of $λ$

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Unification

The Transformation

The Geometry of the Parabola

The Boundaries and Victory

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The set of all values of λ for which the equation cos22x−2sin4x−2cos2x=λ has a solution is

Select Answer:

Visualized Solution

Analyze the Equation

Apply Trigonometric Identities

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } t=cos2x

Expand .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (2t−1)2

Expand .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } −2(1−t)2