A value of x satisfying +3 =2 is:

MCQ

A value of $x$ satisfying $\cos\text{x}+\sqrt{3}\sin\text{x}=2$ is:

A
$\frac{5\pi}{3}$
B
$\frac{4\pi}{3}$
C
$\frac{2\pi}{3}$
✓
$\frac{\pi}{3}$

✓

Answer

Correct option: D.

$\frac{\pi}{3}$

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:
$\text{r}\cos\alpha\cos\text{x}+\text{r}\sin\alpha\sin\text{x}=2$
$\Rightarrow\text{r}\cos(\text{x}-\alpha)=2$
$\Rightarrow\text{r}\cos(\text{x}-\frac{\pi}{3})=2$
$\Rightarrow\text{2}\cos(\text{x}-\frac{\pi}{3})=2$
$\Rightarrow\text{}\cos(\text{x}-\frac{\pi}{3})=1$
$\Rightarrow\text{}\cos(\text{x}-\frac{\pi}{3})=\cos0$
$\Rightarrow\text{x}-\frac{\pi}{3}=0$
$\Rightarrow\text{x}=\frac{\pi}{3}$

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