Trigonometric Functions — MATHS STD 11 Science — Question

1. $\frac{5\pi}{6}$

Solution:

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 $\frac{5\pi}{6}.$