The set of all values of \(λ\) for which the equation \(cos^22x−2sin^4x−2cos^2x=λ\) has a real solution x, is
- \([-1,-\frac{1}{2}]\)
- \([-\frac{3}{2},-1]\)
- \([-\frac{3}{2},-1]\)
- [-2,-1]
The Correct Option is B
Solution and Explanation
λ = cos²2x - 2sin⁴x - 2cosx
Convert all in to cos x.
λ = (2cos²x - 1)² - 2(1 - cos²x)² - 2cosx
= 4cos⁴x - 4cos²x + 1 - 2(1 - 2cos²x + cos⁴x) - 2cosx
= 2cos⁴x - 2cos²x + 1 - 2
= 2cos⁴x - 2cos²x - 1
= 2 [cos⁴x - cos²x - 1/2]
= 2 [(cos²x - 1/2)² - 3/4]
λmax = 2 [1 - 3/4] = 2 × (2/4) = -1 (max value)
λmin = 2 [0 - 3/4] = -3/2 (minimum value)
So, Range = [-3/2, -1]
Top Questions on Trigonometric Equations
If $ \theta \in [-2\pi,\ 2\pi] $, then the number of solutions of $$ 2\sqrt{2} \cos^2\theta + (2 - \sqrt{6}) \cos\theta - \sqrt{3} = 0 $$ is:
- JEE Main - 2025
- Mathematics
- Trigonometric Equations
- If \[ \lim_{x \to 0} \frac{\cos(2x) + a \cos(4x) - b}{x^4} \] is finite, then \( a + b = \) __.
- JEE Main - 2025
- Mathematics
- Trigonometric Equations
- Let $ A = \left\{ \theta \in [0, 2\pi] : \Re\left( \frac{2 \cos \theta + i \sin \theta}{\cos \theta - 3i \sin \theta} \right) = 0 \right\} $. Then $ \sum_{\theta \in A} \theta^2 $ is equal to:
- JEE Main - 2025
- Mathematics
- Trigonometric Equations
- If for $ \theta \in \left[ -\frac{\pi}{3}, 0 \right] $, the points $ (x, y) = \left( 3 \tan\left( \theta + \frac{\pi}{3} \right), 2 \tan\left( \theta + \frac{\pi}{6} \right) \right) $ lie on $ xy + \alpha x + \beta y + \gamma = 0 $, then $ \alpha^2 + \beta^2 + \gamma^2 $ is equal to:
- JEE Main - 2025
- Mathematics
- Trigonometric Equations
- The sum of all values of \( \theta \in [0, 2\pi] \) satisfying \( 2\sin^2\theta = \cos 2\theta \) and \( 2\cos^2\theta = 3\sin\theta \) is:
- JEE Main - 2025
- Mathematics
- Trigonometric Equations
Questions Asked in JEE Main exam
- Find the output voltage in the given circuit.
- JEE Main - 2025
- Current electricity
- Find the equivalent resistance between two ends of the following circuit:
- JEE Main - 2025
- Current electricity
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R):
Assertion (A): In an insulated container, a gas is adiabatically shrunk to half of its initial volume. The temperature of the gas decreases.
Reason (R): Free expansion of an ideal gas is an irreversible and an adiabatic process.In the light of the above statements, choose the correct answer from the options given below:
- JEE Main - 2025
- Current electricity
- A wire of length $ 25 \, \text{m} $ and cross-sectional area $ 5 \, \text{mm}^2 $ having resistivity $ 2 \times 10^{-6} \, \Omega \cdot \text{m} $ is bent into a complete circle. The resistance between diametrically opposite points will be:
- JEE Main - 2025
- Current electricity
Current passing through a wire as function of time is given as $I(t)=0.02 \mathrm{t}+0.01 \mathrm{~A}$. The charge that will flow through the wire from $t=1 \mathrm{~s}$ to $\mathrm{t}=2 \mathrm{~s}$ is:
- JEE Main - 2025
- Current electricity
Concepts Used:
Trigonometric Equations
Trigonometric equation is an equation involving one or more trigonometric ratios of unknown angles. It is expressed as ratios of sine(sin), cosine(cos), tangent(tan), cotangent(cot), secant(sec), cosecant(cosec) angles. For example, cos2 x + 5 sin x = 0 is a trigonometric equation. All possible values which satisfy the given trigonometric equation are called solutions of the given trigonometric equation.
A list of trigonometric equations and their solutions are given below:
| Trigonometrical equations | General Solutions |
| sin θ = 0 | θ = nπ |
| cos θ = 0 | θ = (nπ + π/2) |
| cos θ = 0 | θ = nπ |
| sin θ = 1 | θ = (2nπ + π/2) = (4n+1) π/2 |
| cos θ = 1 | θ = 2nπ |
| sin θ = sin α | θ = nπ + (-1)n α, where α ∈ [-π/2, π/2] |
| cos θ = cos α | θ = 2nπ ± α, where α ∈ (0, π] |
| tan θ = tan α | θ = nπ + α, where α ∈ (-π/2, π/2] |
| sin 2θ = sin 2α | θ = nπ ± α |
| cos 2θ = cos 2α | θ = nπ ± α |
| tan 2θ = tan 2α | θ = nπ ± α |