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Write the value of \theta in [0,2\pi)  where \cos^2 \theta-\sin \theta=\frac{1}{4} 

Option: 1

\{\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6},\frac{11\pi}{6} \}


Option: 2

\{\frac{\pi}{6},\frac{5\pi}{6},\frac{7\pi}{6} \}


Option: 3

\{\frac{\pi}{12},\frac{11\pi}{12} \}


Option: 4

\{\frac{\pi}{6},\frac{5\pi}{6} \}


Answers (1)

best_answer

\\\cos^2 \theta-\sin \theta=\frac{1}{4}\\ 1-\sin^2 \theta-\sin \theta=\frac{1}{4}\\ \sin^2 \theta+\sin \theta-\frac{3}{4}=0\\ 4\sin^2 \theta+4\sin \theta-3=0\\ 4\sin^2 \theta+6\sin \theta-2\sin \theta-3=0\\ (2\sin \theta+3)(2\sin \theta-1)=0\\ \sin \theta=\frac{1}{2}\ \ \ or\ \sin \theta=-\frac{3}{2}(not\ possible )\\ \sin \theta=\sin \frac{\pi}{6}\\ \theta = n\pi+(-1)^n\frac{\pi}{6},\;\;n\in\mathbb{I}\\ \theta=\{\frac{\pi}{6},\frac{5\pi}{6} \}\\

(As only n = 0 and 1 gives answers lying in [0, 2π))

Posted by

Kuldeep Maurya

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