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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Trigonometric Equations

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

The general solution of the system of equations ^ 3 x + ^ 3 ((2...

The general solution of the system of equations \sin^{3} \text{x} + \sin^{3} \left(\dfrac{2 \pi}{3} + \text{x}\right) + \sin^{3} \left(\dfrac{4 \pi}{3} + \text{x}\right) + \dfrac{3}{4} \text\cos 2 \text{x} = 0 and cosx≠0cos x \ne 0 is

Options

  1. Ax=(2k+1)π10, k∈Zx = \dfrac{\left(2\text{k} + 1\right)\pi}{10}\text{, k} \in \text{Z}
  2. Bx=(2k+1)π5, k∈Zx = \dfrac{\left(2\text{k} + 1\right)\pi}{5}\text{, k} \in \text{Z}
  3. Cx=(4k+1)π10, k∈Zx = \dfrac{\left(4\text{k} + 1\right)\pi}{10}\text{, k} \in \text{Z}
  4. Dx=(4k+15)π, k∈Zx = \left(\dfrac{4\text{k} + 1}{5}\right)\pi\text{, k} \in \text{Z}

Correct answer

C. x=(4k+1)π10, k∈Zx = \dfrac{\left(4\text{k} + 1\right)\pi}{10}\text{, k} \in \text{Z}

Step-by-step solution

∵sin⁡3x=3sin⁡x−4sin⁡3x⇒sin⁡3x=14[3sin⁡x−sin⁡3x]∵ \sin 3x = 3\sin x - 4\sin^{3}x ⇒ \sin^{3}x = \dfrac{1}{4}\left[3\sin x - \sin 3x\right]

The given equation reduces to

14(3sin⁡x−sin⁡3x)+14(3sin⁡(2π3+x)−sin⁡(2π+3x))+14(3sin⁡(4π3+x)−sin⁡(4π+3x)+34cos⁡2x=0\dfrac{1}{4}\left(3\sin x - \sin 3x\right) + \dfrac{1}{4}\left(3\sin \left(\dfrac{2\pi}{3} + x\right) - \sin \left(2\pi + 3x\right)\right) + \dfrac{1}{4} ( 3\sin \left(\dfrac{4\pi}{3} + x\right) - \sin \left(4\pi + 3x\right) + \dfrac{3}{4}\cos 2x = 0

3{sinx+sin⁡(2π3+x)+sin⁡(4π3+x)}−{sin⁡3x+sin⁡(2π+3x)+sin⁡(4π+3x)}+3cos⁡2x=03\left\{sin x + \sin \left(\dfrac{2\pi}{3} + x\right) + \sin \left(\dfrac{4\pi}{3} + x\right)\right\} - \left\{\sin 3x + \sin \left(2\pi + 3x\right) + \sin \left(4\pi + 3x\right)\right\} + 3\cos 2x = 0

3{sin⁡x+2sin⁡xcos⁡2π3}−3sin⁡3x+3cos⁡2x=03\left\{\sin x + 2\sin x \cos \dfrac{2\pi}{3}\right\} - 3\sin 3x + 3\cos 2x = 0

⇒sin3x=cos2x⇒cos(π2−3x)=cos2x⇒ \text{sin}3\text{x} = \text{cos}2\text{x} ⇒ \text{cos}\left(\dfrac{\pi}{2} - 3\text{x}\right) = \text{cos}2\text{x}⇒2x=2kπ±(π2−3x)⇒ 2x = 2k \pi \pm \left(\dfrac{\pi}{2} - 3x\right)

⇒5x=2kπ+π2=(4k+1)π2⇒ 5\text{x} = 2\text{k}\pi + \dfrac{\pi}{2} = \left(4\text{k} + 1\right)\dfrac{\pi}{2} ∀k∈Z\forall k \in Z or x=−2kπ+π2,∀k∈Zx =-2 k \pi + \dfrac{\pi}{2}, \forall k \in Z

⇒x=(4k+1)π10,∀k∈Z⇒ \text{x} = \left(4\text{k} + 1\right)\dfrac{\pi}{10}\text{,} \forall k \in Z{cosx≠0}\left\{cos x \ne 0\right\}
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