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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
If $\cos\text{x}+\sqrt{3}\sin\text{x}=2,\text{then}\ \text{x}=$
  • ✓
    $\frac{\pi}{3}$
  • B
    $\frac{2\pi}{3}$
  • C
    $\frac{4\pi}{3}$
  • D
    $\frac{5\pi}{3}$
✓

Answer

Correct option: A.
$\frac{\pi}{3}$
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 $\text{n}=0,\text{x}=\frac{\pi}{3}.$
$\therefore\text{x}=\frac{\pi}{3}$

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