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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
Solve the following equations:
$\sqrt{3}\cos\text{x}+\sin\text{x}=1$
✓

Answer

We have,

$\sqrt{3}\cos\text{x}+\sin\text{x}=1$

Divide both side by 2, we get

$\frac{\sqrt{3}}{2}\cos\text{x}+\frac{1}{2}\sin\text{x}=\frac{1}{2}$

$\Rightarrow\cos\frac{\pi}{6}\cos\text{x}+\sin\frac{\pi}{6}\sin\text{x}=\frac{1}{2}$ $\Big[\because\sin\frac{\pi}{6}=\frac{1}{2},\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}\Big]$

$\Rightarrow\cos\Big(\text{x}-\frac{\pi}{6}\Big)=\cos\frac{\pi}{3}$

$\Rightarrow\text{x}=\frac{\pi}{6}=2\text{n}\pm\frac{\pi}{3},\text{n}\in\text{z}$

$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3}+\frac{\pi}{6},\text{n}\in\text{z}$

$\Rightarrow\text{x}=(4\text{n}+1)\frac{\pi}{2}$ or $(12\text{m}-1)\frac{\pi}{6},\text{n},\text{m}\in\text{z}$

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