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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
A solution of the equation $\cos^2\text{x}+\sin\text{x}+1=0,$ lies in the interval:
  • A
    $\Big(\frac{-\pi}{4},\ \frac{\pi}{4}\Big)$
  • B
    $\Big(\frac{\pi}{4},\ \frac{3\pi}{4}\Big)$
  • C
    $\Big(\frac{3\pi}{4},\ \frac{5\pi}{4}\Big)$
  • ✓
    $\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big)$
✓

Answer

Correct option: D.
$\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big)$
Given:
$\cos^2\text{x}+\sin\text{x}+1=0$
$\Rightarrow(1-\sin^2\text{x})+\sin\text{x}+1=0$
$\Rightarrow1-\sin^2\text{x}+\sin\text{x}+1=0$
$\Rightarrow\sin^2\text{x}-\sin\text{x}-2=0$
$\Rightarrow\sin^2\text{x}-2\sin\text{x}+\sin\text{x}-2=0$
$\Rightarrow\sin\text{x}(\sin\text{x}-2)+1(\sin\text{x}-2)=0$
$\Rightarrow(\sin\text{x}-2)(\sin\text{x}+1)=0$
$\Rightarrow\sin\text{x}-2=0$ or $\sin\text{x}+1=0$
$\Rightarrow\sin\text{x}=2$ or $\sin\text{x}=-1$
$\sin\text{x}=2$ is not possible.
$\Rightarrow\sin\text{x}=-1$
$\therefore\sin\text{x}=\sin\frac{3\pi}{2}$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{3\pi}{2},\ \text{n}\in\text{Z}$
The values of $x$ lies in the third and fourth quadrants.
Hence, $x$ lies in $\Big(\frac{5\pi}{4},\ \frac{7\pi}{4}\Big).$

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