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Trigonometrical Equations
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સમીકરણ ${\cos ^2}\theta + \sin \theta + 1 = 0$ નો ઉકેલ . . . . અંતરાલમાં આવેલ છે.
A
$\left( { - \frac{\pi }{4},\frac{\pi }{4}} \right)$
B
$\left( {\frac{\pi }{4},\frac{{3\pi }}{4}} \right)$
C
$\left( {\frac{{3\pi }}{4},\frac{{5\pi }}{4}} \right)$
D
$\left( {\frac{{5\pi }}{4},\frac{{7\pi }}{4}} \right)$
(IIT-1992)
Solution
(d) We have, ${\cos ^2}\theta + \sin \theta + 1 = 0$
==> $1 – {\sin ^2}\theta + \sin \theta + 1 = 0$
==> ${\sin ^2}\theta – \sin \theta – 2 = 0$
==> $(\sin \theta + 1)\,(\sin \theta – 2) = 0$
$\sin \theta = 2$, which is not possible and $\sin \theta = – 1$.
Therefore, solution of given equation lies in the interval $\left( {\frac{{5\pi }}{4},\,\frac{{7\pi }}{4}} \right)$.
Standard 11
Mathematics
