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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
Solve the following equations:
$\sin\text{x}-3\sin2\text{x}+\sin3\text{x}=\cos\text{x}-3\cos2\text{x}+\cos3\text{x}$
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Answer

$\sin\text{x}-3\sin2\text{x}+\sin3\text{x}=\cos\text{x}-3\cos2\text{x}+\cos3\text{x}$
$(\sin\text{x}+\sin3\text{x})-3\sin2\text{x}-(\cos\text{x}+\cos3\text{x})+3\cos2\text{x}=0$
$2\sin2\text{x}\cos\text{x}-3\sin2\text{x}-2\cos2\text{x}\cos\text{x}+3\cos2\text{x}=0$
$\sin2\text{x}(2\cos\text{x}-3)-\cos2\text{x}(2\cos\text{x}-3)=0$
$(2\cos\text{x}-3)(\sin2\text{x}-\cos2\text{x})=0$
$\cos\text{x}=\frac{3}{2}$ or $\sin2\text{x}-\cos2\text{x}-\cos2\text{x}=0$
but $\cos\text{x}\in[-11]\Rightarrow\cos\text{x}\not=\frac{3}{2}$
$\sin2\text{x}=\cos2\text{x}$
$2\text{x}=\text{n}\pi+\frac{\pi}{4}$
$\text{x}=\frac{\text{n}\pi}{2}+\frac{\pi}{8}$

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