The number of solutions of the equation ( 4 – 3 ) sin x – 2 3 cos 2 x = – 4 1 + 3 , x [ – 2 , 5 2 ] is [2025]

Question

The number of solutions of the equation $(4 - \sqrt{3}) \sin x - 2 \sqrt{3} \cos^{2} x = - \frac{4}{1 + \sqrt{3}}, x \in [- 2 π, \frac{5 π}{2}]$ is [2025]

Smart Achievers · Accepted Answer

(1)

Given, $(4 - \sqrt{3}) \sin x - 2 \sqrt{3} \cos^{2} x = - \frac{4}{1 + \sqrt{3}}, x \in [- 2 π, \frac{5 π}{2}]$

$\Rightarrow (4 - \sqrt{3}) \sin x - 2 \sqrt{3} (1 - \sin^{2} x) = 2 (1 - \sqrt{3})$

$\Rightarrow 2 \sqrt{3} \sin^{2} x + 4 \sin x - \sqrt{3} s in x - 2 = 0$

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

$\Rightarrow \sin x = \frac{1}{2} [∵ \sin x \neq \frac{- 2}{\sqrt{3}}]$

$\Rightarrow x = \frac{- 11 π}{6}, \frac{- 7 π}{6}, \frac{π}{6}, \frac{5 π}{6}, \frac{13 π}{6}$

$∴$ Number of solutions = 5.

Solution

Q.
The number of solutions of the equation $(4 - \sqrt{3}) \sin x - 2 \sqrt{3} \cos^{2} x = - \frac{4}{1 + \sqrt{3}}, x \in [- 2 π, \frac{5 π}{2}]$ is [2025]

1 5

2 3

3 6

4 4

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Solution

Q. The number of solutions of the equation (4–3)sinx–23cos2x=–41+3,x∈[–2π,5π2] is [2025]

1 5

2 3

3 6

4 4

Ans. (1) Given, (4–3)sinx–23cos2x=–41+3, x∈[–2π,5π2] ⇒ (4–3)sinx–23(1–sin2x)=2(1–3) ⇒ 23sin2x+4sinx–3sinx–2=0 ⇒ (2sinx–1)(3sinx+2)=0 ⇒ sinx=12 [∵ sinx≠–23] ⇒ x=–11π6,–7π6,π6,5π6,13π6 ∴ Number of solutions = 5.

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Q.
The number of solutions of the equation $(4 - \sqrt{3}) \sin x - 2 \sqrt{3} \cos^{2} x = - \frac{4}{1 + \sqrt{3}}, x \in [- 2 π, \frac{5 π}{2}]$ is [2025]