If θ ∈ [-7π/6, 4π/3], then the number of solutions of √3 cosec²θ - 2(√3-1)cosecθ - 4 = 0, is equal to :

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

Maths Q13 – JEE-MAIN 2025

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Home Maths Trigonometric FunctionsQuestion #490

θ \in [- \frac{7 π}{6}, \frac{4 π}{3}]

, then the number of solutions of

\sqrt{3} {cosec}^{2} θ - 2 (\sqrt{3} - 1) cosec θ - 4 = 0

, is equal to :

Clue 1 of 3: Quadratic Equation in Trigonometric Ratios

Recognize the given equation as a quadratic equation in terms of $cosec θ$ .

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Step 1: Solve the quadratic equation for

cosec θ

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The given equation is $\sqrt{3} {cosec}^{2} θ - 2 (\sqrt{3} - 1) cosec θ - 4 = 0$ . Let $x = cosec θ$ . The equation becomes $\sqrt{3} x^{2} - 2 (\sqrt{3} - 1) x - 4 = 0$ . Using the quadratic formula $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ :

x = \frac{2 (\sqrt{3} - 1) \pm \sqrt{4 (\sqrt{3} - 1)^{2} - 4 (\sqrt{3}) (- 4)}}{2 \sqrt{3}}

x = \frac{2 (\sqrt{3} - 1) \pm \sqrt{4 (3 - 2 \sqrt{3} + 1) + 16 \sqrt{3}}}{2 \sqrt{3}}

x = \frac{2 (\sqrt{3} - 1) \pm \sqrt{16 - 8 \sqrt{3} + 16 \sqrt{3}}}{2 \sqrt{3}}

x = \frac{2 (\sqrt{3} - 1) \pm \sqrt{16 + 8 \sqrt{3}}}{2 \sqrt{3}}

x = \frac{2 (\sqrt{3} - 1) \pm 2 \sqrt{(\sqrt{3} + 1)^{2}}}{2 \sqrt{3}}

x = \frac{(\sqrt{3} - 1) \pm (\sqrt{3} + 1)}{\sqrt{3}}

This yields two solutions: $x_{1} = \frac{\sqrt{3} - 1 + \sqrt{3} + 1}{\sqrt{3}} = \frac{2 \sqrt{3}}{\sqrt{3}} = 2$ and $x_{2} = \frac{\sqrt{3} - 1 - (\sqrt{3} + 1)}{\sqrt{3}} = \frac{- 2}{\sqrt{3}}$ . Therefore, $cosec θ = 2$ or $cosec θ = - \frac{2}{\sqrt{3}}$ . This implies $\sin θ = \frac{1}{2}$ or $\sin θ = - \frac{\sqrt{3}}{2}$ .

Step 2: Find solutions for

\sin θ = \frac{1}{2}

in the given interval○ Expand

The given interval is $θ \in [- \frac{7 π}{6}, \frac{4 π}{3}]$ , which corresponds to $[- 210^{\circ}, 240^{\circ}]$ . For $\sin θ = \frac{1}{2}$ , the general solutions are $θ = n π + (- 1)^{n} \frac{π}{6}$ . Checking values within the interval:

- For $n = 0$ , $θ = \frac{π}{6}$ ( $30^{\circ}$ ). This is in the interval. - For $n = 1$ , $θ = π - \frac{π}{6} = \frac{5 π}{6}$ ( $150^{\circ}$ ). This is in the interval. - For $n = - 1$ , $θ = - π - \frac{π}{6} = - \frac{7 π}{6}$ ( $- 210^{\circ}$ ). This is in the interval (boundary).

Solutions for $\sin θ = \frac{1}{2}$ in the interval are ${- \frac{7 π}{6}, \frac{π}{6}, \frac{5 π}{6}}$ . (3 solutions)

💡 Teacher's Secret Hint

Remember to include boundary values if they satisfy the equation.

Step 3: Find solutions for

\sin θ = - \frac{\sqrt{3}}{2}

in the given interval and count total solutions○ Expand

For $\sin θ = - \frac{\sqrt{3}}{2}$ , the general solutions are $θ = 2 n π - \frac{π}{3}$ or $θ = 2 n π + \frac{4 π}{3}$ . Checking values within the interval:

- For $n = 0$ , $θ = - \frac{π}{3}$ ( $- 60^{\circ}$ ). This is in the interval. - For $n = 0$ , $θ = \frac{4 π}{3}$ ( $240^{\circ}$ ). This is in the interval (boundary). - For $n = - 1$ , $θ = - 2 π + \frac{4 π}{3} = - \frac{2 π}{3}$ ( $- 120^{\circ}$ ). This is in the interval.

Solutions for $\sin θ = - \frac{\sqrt{3}}{2}$ in the interval are ${- \frac{2 π}{3}, - \frac{π}{3}, \frac{4 π}{3}}$ . (3 solutions) All 6 solutions are distinct. The total number of distinct solutions is $3 + 3 = 6$ .

💡 Teacher's Secret Hint

Carefully check all quadrants and the periodicity of sine function within the specified range.

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Maths Question 13 – JEE-MAIN 2025

Clue 1 of 3: Quadratic Equation in Trigonometric Ratios

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