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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Mathematics
Chapter
Trigonometric Equations

BITSAT 2014 · Mathematics · Trigonometric Equations

The solution of (2 x-1)(3+2 x)=0 in the interval 0 ≤ x ≤ 2 π is

The solution of (2cos⁡x−1)(3+2cos⁡x)=0(2 \cos x-1)(3+2 \cos x)=0 in the interval 0≤x≤2π0 \leq x \leq 2 \pi is

Options

  1. Aπ3\frac{\pi}{3}
  2. Bπ3,5π3\frac{\pi}{3}, \frac{5 \pi}{3}
  3. Cπ3,5π3,cos⁡−1(−32)\frac{\pi}{3}, \frac{5 \pi}{3}, \cos ^{-1} \left(-\frac{3}{2}\right)
  4. DNone of these

Correct answer

B. π3,5π3\frac{\pi}{3}, \frac{5 \pi}{3}

Step-by-step solution

We have (2cos⁡x−1)(3+2cos⁡x)=0(2 \cos x-1)(3+2 \cos x)=0 If 2cos⁡x−1=0,2 \cos x-1=0, then cos⁡x=12∴x=π/3,5π/3\cos x=\frac{1}{2}\therefore x=\pi / 3,5 \pi / 3 If 3+2cos⁡x=0,3+2 \cos x=0, the cos⁡x=−3/2\cos x=-3 / 2 which is not possible.
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