MCQ
The smallest positive angle which satisfies the equation $2\sin^2\text{x}+\sqrt{3}\cos\text{x}+1=0$ is:
- ✓$\frac{5\pi}{6}$
- B$\frac{2\pi}{3}$
- C$\frac{\pi}{3}$
- D$\frac{\pi}{6}$
✓
Answer
Correct option: A.
$\frac{5\pi}{6}$
Given:
$2\sin^2\text{x}+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2(1-\cos^2\text{x})+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2-2\cos^2\text{x}+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2\cos^2\text{x}-\sqrt{3}\cos\text{x}-3=0$
$\Rightarrow2\cos^2\text{x}-2\sqrt{3}\cos\text{x}+\sqrt{3}\cos\text{x}-3=0$
$\Rightarrow2\cos\text{x}\Big(\cos\text{x}-\sqrt{3}\Big)+\sqrt{3}\Big(\cos\text{x}-\sqrt{3}\Big)=0$
$\Rightarrow\Big(2\cos\text{x}+\sqrt{3}\Big)\Big(\cos\text{x}-\sqrt{3}\Big)=0$
$\therefore\cos\text{x}+\sqrt{3}=0$ or, $\cos\text{x}=\sqrt{3}$ is not possible.
$\Rightarrow\cos\text{x}=\cos\Big(\frac{\pi}{6}\Big)$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{5\pi}{6},\ \text{n}\in\text{Z}$
For $n = 0$, the value of $x$ is $\pm\frac{5\pi}{6}.$
Hence, the smallest positive angle is $\frac{5\pi}{6}.$
$2\sin^2\text{x}+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2(1-\cos^2\text{x})+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2-2\cos^2\text{x}+\sqrt{3}\cos\text{x}+1=0$
$\Rightarrow2\cos^2\text{x}-\sqrt{3}\cos\text{x}-3=0$
$\Rightarrow2\cos^2\text{x}-2\sqrt{3}\cos\text{x}+\sqrt{3}\cos\text{x}-3=0$
$\Rightarrow2\cos\text{x}\Big(\cos\text{x}-\sqrt{3}\Big)+\sqrt{3}\Big(\cos\text{x}-\sqrt{3}\Big)=0$
$\Rightarrow\Big(2\cos\text{x}+\sqrt{3}\Big)\Big(\cos\text{x}-\sqrt{3}\Big)=0$
$\therefore\cos\text{x}+\sqrt{3}=0$ or, $\cos\text{x}=\sqrt{3}$ is not possible.
$\Rightarrow\cos\text{x}=\cos\Big(\frac{\pi}{6}\Big)$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{5\pi}{6},\ \text{n}\in\text{Z}$
For $n = 0$, the value of $x$ is $\pm\frac{5\pi}{6}.$
Hence, the smallest positive angle is $\frac{5\pi}{6}.$
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