Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
Solve the following equation: $2\cos^{2}\text{x}-5\cos\text{x}+2=0$
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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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