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Trigonometric Equations — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Equations5 Marks
Question
Solve the following equations:
$3\sin2\text{x}-5 \sin\text{x}\cos \text{x} + 8 \cos2\text{x = 2}$
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Answer

$3\sin^2\text{x}-5\sin\text{x}\cos\text{x}+8\cos^2\text{x}=2$
$\Rightarrow3\sin^2\text{x}-5\sin\text{x}\cos\text{x}+3\cos^\text{x}2+5\cos^2\text{x}-2=0$
$\Rightarrow3(\sin^2\text{x}\cos^2\text{x})-5\sin\text{x}\cos\text{x}+5\cos^2\text{x}-2=0$
$\Rightarrow3-5\sin\text{x}\cos\text{x}+5\cos^2\text{x}-2=0$
$\Rightarrow5\cos^2\text{x}-5\sin\text{x}\cos\text{x}+1=0$
$\Rightarrow5(1-\sin^2\text{x})-5\sin\text{x}\cos\text{x}+1=0$
$\Rightarrow5-5\sin^2\text{x}-5\sin\text{x}\cos\text{x}+1=0$
$\Rightarrow5\sin^2\text{x}+5\sin\text{x}\cos\text{x}-6=0$
Dividing by $\cos^2\text{x},\ $ we get
$\Rightarrow5\tan^2\text{x}+5\tan\text{x}-6\sec^2\text{x}=0$
$\Rightarrow5\tan^2\text{x}+5\tan\text{x}-6-6\tan^2\text{x}=0$
$\Rightarrow-\tan^2\text{x}+5\tan\text{x}-6=0$
$\Rightarrow\tan^2\text{x}-5\tan\text{x}+6=0$
$\Rightarrow\tan^2\text{x}-3\tan\text{x}-2\tan\text{x}+6=0$
$\Rightarrow(\tan\text{x}-3)=0$ or $\tan\text{x}=2$
$\Rightarrow\text{x}=\text{n}\pi+\tan^{-1}3$ or $\text{x}=\text{n}\pi+\tan6{-1}2,\ \text{n}\in\text{Z}$

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