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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
Solve the following equations:
$3-2\cos\text{x}-4\sin\text{x}-\cos2\text{x}+\sin2\text{x}=0$
✓

Answer

$3-2\cos\text{x}-4\sin\text{x}-\cos2\text{x}+\sin2\text{x}=0$
$\Rightarrow3-2\cos\text{x}-4\sin\text{x}-(1-2\sin^2\text{x})+2\sin\text{x}\cos\text{x}=0$
$\Rightarrow3-2\cos\text{x}-4\sin\text{x}-1+2\sin^2\text{x}+2\sin\text{x}\cos\text{x}=0$
$\Rightarrow(2\sin^2\text{x}-4\sin\text{x}+2)+2\cos\text{x}(\sin\text{x}-1)=0$
$\Rightarrow2(\sin^2\text{x}-2\sin\text{x}+1)+2\cos\text{x}(\sin\text{x}-1)=0$
$\Rightarrow2(\sin\text{x}-1)^2+2\cos\text{x}(\sin\text{x}-1)=0$
$\Rightarrow(\sin\text{x}-1)(2\sin\text{x}-2+2\cos\text{x})=0$
$\Rightarrow2(\sin\text{x}-1)(\sin\text{x}+\cos\text{x}-1)=0$
$\Rightarrow(\sin\text{x}-1)=0$ or $(\sin\text{x})+\cos\text{x}-1=0$
$\Rightarrow\sin\text{x}=1$ or $\sin\text{x}+\cos\text{x}=1$
$\Rightarrow\sin\text{x}=\sin\frac{\pi}{2}$ or $\frac{1}{\sqrt{2}}\sin\text{x}+\frac{1}{\sqrt{2}}\cos\text{x}=\frac{1}{\sqrt{2}}$
$\Rightarrow\sin\text{x}=\sin\frac{\pi}{2}$ or $\sin\frac{\pi}{4}\sin\text{x}+\cos\frac{\pi}{4}\cos\text{x}=\cos\frac{\pi}{4}$
$\Rightarrow\sin\text{x}=\sin\frac{\pi}{2}$ or $\cos\Big(\text{x}-\frac{\pi}{4}\Big)=\cos\frac{\pi}{4}$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{2}$ or $\text{x}-\frac{\pi}{4}=2\text{n}\pi\pm\frac{\pi}{4},\ \text{n}\in\text{Z}$
$\Rightarrow\text{x}=\text{n}\pi+(-1)^\text{n}$ or $\text{x}=2\text{n}\pi+\frac{\pi}{2}$ or $\text{x}=2\text{n}\pi,\ 2\text{n}\pi,\ \text{n}\in\text{Z}$

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