CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Equations5 Marks
Question
Solve the following equation:
$2\cos^{2}\text{x}-5\cos\text{x}+2=0$
✓
Answer
We have,
$2\cos^{2}\text{x}-5\cos\text{x}+2=0$
$\Rightarrow2\cos^{2}\text{x}-4\cos\text{x}-\cos\text{x}+2=0$ [Use factorization]
$\Rightarrow2\cos\text{x}(\cos\text{x}-2)-1(\cos\text{x}-2)=0$
$\Rightarrow(2\cos\text{x}-1)(\cos\text{x}-2)=0$
$\Rightarrow\text{Either}$
$2\cos\text{x}-1=0$ or $\cos\text{x}-2=0$
$\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=2$
$\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$ $\big[$This is not possible as $-1<\cos\text{x}<1\big]$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$
Thus,
$\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$
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