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Trigonometric Equations — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Equations3 Marks
Question
Solve the following equations:
$\sin3\text{x}-\sin\text{x}=4\cos^{2}\text{x}-2$
✓

Answer

$\sin3\text{x}-\sin\text{x}=4\cos^{2}\text{x}-2$
$\Rightarrow2\cos2\text{x},\sin\text{x}=2(2\cos^{2}\text{x}-1)$
$\Rightarrow2\cos2\text{x},\sin\text{x}=2\cos2\text{x}$ $[\because\cos2\text{x}=2\cos^{2}\text{x}-1]$
$\Rightarrow2\cos2\text{x}(\sin\text{x}-1)=0$
Either
$\cos2\text{x}=0$ or $\sin\text{x}-1=0$
$\Rightarrow2\text{x}=(2\text{n}+1)\frac{\pi}{2},\text{n}\in\text{z}$or $\sin\text{x}=1=\sin\frac{\pi}{2}$
$\Rightarrow\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\text{x}=\text{m}\pi+(-1)^\text{m}\frac{\pi}{2},\text{m}\in\text{z}$
Thus,
$\text{x}=(2\text{n}+1)\frac{\pi}{4},\text{n}\in\text{z}$ or $\text{m}\pi+(-1)^\text{m}\frac{\pi}{2},\text{m}\in\text{z}$

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