If θ ∈ [-2π, 2π], then the number of solutions of 2√2 cos²θ + (2-√6)cosθ - √3=0, is equal to :

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q3 – JEE-MAIN 2025

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θ \in [- 2 π, 2 π]

, then the number of solutions of

2 \sqrt{2} \cos^{2} θ + (2 - \sqrt{6}) \cos θ - \sqrt{3} = 0

, is equal to :

Clue 1 of 3: Quadratic Form Recognition

The given equation can be treated as a quadratic equation by substituting a variable for $\cos θ$ .

🧠 Full Solution Path

● Video Walkthrough

Step 1: Transform the equation into a quadratic form and solve for

\cos θ

.✦ Active

Let $x = \cos θ$ . The given equation $2 \sqrt{2} \cos^{2} θ + (2 - \sqrt{6}) \cos θ - \sqrt{3} = 0$ transforms into a quadratic equation in $x$ :

2 \sqrt{2} x^{2} + (2 - \sqrt{6}) x - \sqrt{3} = 0

We can factor this quadratic equation by splitting the middle term. We look for two numbers whose product is $(2 \sqrt{2}) (- \sqrt{3}) = - 2 \sqrt{6}$ and whose sum is $(2 - \sqrt{6})$ . These numbers are $2$ and $- \sqrt{6}$ .

2 \sqrt{2} x^{2} + 2 x - \sqrt{6} x - \sqrt{3} = 0

Factor by grouping:

2 x (\sqrt{2} x + 1) - \sqrt{3} (\sqrt{2} x + 1) = 0

(\sqrt{2} x + 1) (2 x - \sqrt{3}) = 0

This gives two possible values for $x = \cos θ$ :

\sqrt{2} x + 1 = 0 ⟹ x = - \frac{1}{\sqrt{2}}

2 x - \sqrt{3} = 0 ⟹ x = \frac{\sqrt{3}}{2}

Step 2: Find the number of solutions for

θ

in the given interval for each value of

\cos θ

.○ Expand

The given interval for $θ$ is $[- 2 π, 2 π]$ , which has a length of $4 π$ . For any value $k \in (- 1, 1)$ , the equation $\cos θ = k$ has two solutions in any interval of length $2 π$ (e.g., $[0, 2 π)$ ). Since the interval $[- 2 π, 2 π]$ covers two such $2 π$ cycles, there will be $2 \times 2 = 4$ solutions for each distinct value of $k$ in $(- 1, 1)$ .

Case 1: $\cos θ = - \frac{1}{\sqrt{2}}$

The principal values are $θ = \frac{3 π}{4}$ and $θ = \frac{5 π}{4}$ . In the interval $[- 2 π, 2 π]$ , the solutions are $\frac{3 π}{4}, \frac{5 π}{4}, - \frac{3 π}{4}, - \frac{5 π}{4}$ . Thus, there are 4 solutions.

Case 2: $\cos θ = \frac{\sqrt{3}}{2}$

The principal values are $θ = \frac{π}{6}$ and $θ = \frac{11 π}{6}$ . In the interval $[- 2 π, 2 π]$ , the solutions are $\frac{π}{6}, \frac{11 π}{6}, - \frac{π}{6}, - \frac{11 π}{6}$ . Thus, there are 4 solutions.

💡 Teacher's Secret Hint

Remember that $\cos θ = k$ has 2 solutions in $[0, 2 π)$ for $k \in (- 1, 1)$ . The interval $[- 2 π, 2 π]$ covers two such cycles.

Step 3: Calculate the total number of solutions.○ Expand

All 8 solutions obtained from Case 1 and Case 2 are distinct. Therefore, the total number of solutions for the given equation in the interval $θ \in [- 2 π, 2 π]$ is the sum of solutions from both cases:

Total solutions = 4 + 4 = 8

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Maths Question 3 – JEE-MAIN 2025

Clue 1 of 3: Quadratic Form Recognition

🧠 Full Solution Path