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

NTA Abhyas 2020 · Mathematics · Trigonometric Equations

Number of roots of the equation ^ 2 x + 3 + 1 2 sin x - 3 4 - 1 = 0...

Number of roots of the equation cos⁡2x+3+12sinx−34−1=0\cos^{2} x + \dfrac{\sqrt{3} + 1}{2}sin x - \dfrac{\sqrt{3}}{4} - 1 = 0 which lie in the interval [−π,π]\left[- \pi, \pi\right] is

Options

  1. A2
  2. B4
  3. C6
  4. D8

Correct answer

B. 4

Step-by-step solution

Given equation is

1−sin⁡2x+3+12sinx−34−1=01 - \sin^{2} x + \dfrac{\sqrt{3} + 1}{2}sin x - \dfrac{\sqrt{3}}{4} - 1 = 0

⇒sin⁡2x−3+12sinx+34=0;4sin⁡2x−23sinx−2sinx+3=0⇒ \sin^{2} x - \dfrac{\sqrt{3} + 1}{2}sin x + \dfrac{\sqrt{3}}{4} = 0; 4\sin^{2} x - 2\sqrt{3}sin x - 2sin x + \sqrt{3} = 0

2sin⁡x(2sin⁡x−3)−(2sin⁡x−3)=02\sin x( 2\sin x - \sqrt{3} ) - ( 2\sin x - \sqrt{3} ) = 0

⇒(2sin⁡x−1)(2sin⁡x−3)=0⇒ ( 2\sin x - 1 )( 2\sin x - \sqrt{3} ) = 0

On solving we get sinx=12;32sin x = \dfrac{1}{2}; \dfrac{\sqrt{3}}{2}

x=π6,5π6;π3,2π3x = \dfrac{\pi}{6}, \dfrac{5\pi}{6}; \dfrac{\pi}{3}, \dfrac{2\pi}{3}
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