Trigonometric Functions — MATHS STD 11 Science — Question

4.  $\frac{\pi}{3}$

Solution:

Given equation:

$\cot\text{x}+\sqrt{3}\sin​​​​\text{x}=2\ .....(1)$

Thus, the 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 $\text{c}=3$

Let:

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

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

$\Rightarrow\text{r}\sqrt{\text{a}^2+\text{b}^2}=\sqrt{\Big(\sqrt{3}\Big)^2+1^2}=2$ and

$\tan\alpha=\frac{\text{b}}{\text{a}}\Rightarrow\tan\alpha=\frac{\sqrt{3}}{1}\Rightarrow\tan\alpha\tan\frac{\pi}{3}\Rightarrow\alpha=\frac{\pi}{3}$

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

$\text{r}\cos\alpha\cos\text{x}+\text{r}\sin\alpha\sin\text{x}=2$

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

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

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

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

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

$\Rightarrow​\text{x}​-\frac{\pi}{3}=0$

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