If [ – 7 6 , 4 3 ] , then the number of solutions of 3 cosec 2 – 2 ( 3 – 1 ) cosec – 4 = 0 , is equal to : [2025]

Question

If $θ \in [- \frac{7 π}{6}, \frac{4 π}{3}]$ , then the number of solutions of $\sqrt{3} {cosec}^{2} θ - 2 (\sqrt{3} - 1) cosec θ - 4 = 0$ , is equal to : [2025]

Smart Achievers · Accepted Answer

(4)

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

Let $cosec θ = x$

$∴ \sqrt{3} x^{2} - 2 (\sqrt{3} - 1) x - 4 = 0$

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

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

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

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

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

$∴ x_{1} = \frac{(\sqrt{3} - 1) + (\sqrt{3} + 1)}{\sqrt{3}} = 2$

and $x_{2} = \frac{(\sqrt{3} - 1) - (\sqrt{3} + 1)}{\sqrt{3}} = \frac{- 2}{\sqrt{3}}$

Now, Put x = $cosec θ$

When $cosec θ = 2 \Rightarrow \sin θ = \frac{1}{2} \Rightarrow θ = \frac{π}{6}, \frac{5 π}{6}, \frac{- 7 π}{6}$

When $cosec θ = \frac{- 2}{\sqrt{3}} \Rightarrow \sin θ = - \frac{\sqrt{3}}{2}$

$\Rightarrow θ = \frac{4 π}{3}, \frac{5 π}{3}, \frac{- π}{3}, \frac{- 2 π}{3}$

Now, $\frac{5 π}{3} \notin [\frac{- 7 π}{6}, \frac{4 π}{3}]$

$∴$ Required number of solutions are = 6.

Solution

Q.
If $θ \in [- \frac{7 π}{6}, \frac{4 π}{3}]$ , then the number of solutions of $\sqrt{3} {cosec}^{2} θ - 2 (\sqrt{3} - 1) cosec θ - 4 = 0$ , is equal to : [2025]

1 7

2 10

3 8

4 6

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Solution

Q. If θ∈[–7π6,4π3], then the number of solutions of 3cosec2θ–2(3–1)cosecθ–4=0, is equal to : [2025]

1 7

2 10

3 8

4 6

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Q.
If $θ \in [- \frac{7 π}{6}, \frac{4 π}{3}]$ , then the number of solutions of $\sqrt{3} {cosec}^{2} θ - 2 (\sqrt{3} - 1) cosec θ - 4 = 0$ , is equal to : [2025]