Trigonometric Functions — MATHS STD 11 Science — Question

1. $\frac{\pi}{3}$

Solution:

Given: $\cos\text{x}+\sqrt{3}\sin\text{x}=2\ ......(1)$

This equation is of the form $\text{a}\cos\text{x}+\text{b}\sin\text{x}=\text{c},$ where $\text{a}=1,\ \text{b}=\sqrt{3}$ and $\text{c}=2.$

Let:

$\text{a}=\text{r}\ \cos\alpha$ and $\text{b}=\sin\alpha$

Now,

$1=\text{r}\ \cos\alpha,\sqrt{3}=\text{r}\sin\alpha$

$\Rightarrow\text{r}=\sqrt{\text{a}^2+\text{b}^2}=\sqrt{1+3}=\sqrt{4}=2$

And,

$\tan\alpha=\frac{\text{b}}{\text{a}}$

$\Rightarrow\tan\alpha=\frac{\sqrt{3}}{1}$

$\Rightarrow\tan\alpha\sqrt{3}$

$\Rightarrow\alpha=\frac{\pi}{3}$

On putting $\text{a}=1=\text{r}\ \cos\alpha$ and $\text{b}=\sqrt{3}=r\ \sin\alpha$ in equation (1), we get:

$\Rightarrow\text{r}\cos(\text{x}-\alpha)=2$

$\Rightarrow2\cos(\text{x}-\frac{\pi}{3})=2$

$\Rightarrow\cos(\text{x}-\frac{\pi}{3})=1$

$\Rightarrow\cos(\text{x}-\frac{\pi}{3})=\cos0$

$\Rightarrow\text{x}=\frac{\pi}{3}=2\text{n}\pi\pm0$

$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3}$

For $\text{n}=0,\text{x}=\frac{\pi}{3}.$

$\therefore\text{x}=\frac{\pi}{3}$