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EEJ MAIN Mathematics QUESTION #979
Question 1
The sum of all values of \(\theta\in[0,2\pi]\) satisfying \(2\sin^2\theta=\cos 2\theta\) and \(2\cos^2\theta=3\sin\theta\) is:
  • \(4\pi\)
  • \(\dfrac{5\pi}{6}\)
  • \(\pi\)✔️
  • \(\dfrac{\pi}{2}\)
Correct Answer Explanation
From \(2\sin^2\theta=\cos2\theta=1-2\sin^2\theta \Rightarrow 4\sin^2\theta=1 \Rightarrow \sin\theta=\pm\frac{1}{2}\). From \(2\cos^2\theta=3\sin\theta \Rightarrow 2(1-\sin^2\theta)=3\sin\theta \Rightarrow 2\sin^2\theta+3\sin\theta-2=0 \Rightarrow \sin\theta=\frac{1}{2}\) (taking positive root). So \(\theta=\frac{\pi}{6}\) or \(\frac{5\pi}{6}\). Sum\(=\frac{\pi}{6}+\frac{5\pi}{6}=\pi\).