Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Equations5 Marks
Question
Solve the following equation: $\sin^{2}\text{x}-\cos\text{x}=\frac{1}{4}$
✓
Answer
We have, $\sin^{2}\text{x}-\cos\text{x}=\frac{1}{4}$ $\Rightarrow1-\cos^{2}\text{x}-\cos\text{x}=\frac{1}{4}$$[\because\sin^{2}\text{x}=1-\cos^{2}\text{x}]$ $\Rightarrow\cos^{2}\text{x}+\cos\text{x}-\frac{3}{4}=0$ $\Rightarrow4\cos^{2}\text{x}+4\cos\text{x}-3=0$ $\Rightarrow4\cos^{2}\text{x}+6\cos\text{x}+2\cos\text{x}-3=0$ [factorize it] $\Rightarrow2\cos\text{x}(2\cos\text{x}+3)-1(\cos\text{x}+3)=0$ $\Rightarrow(2\cos\text{x}-1)(2\cos\text{x}+3)=0$ $\Rightarrow\text{ Either}$ $2\cos\text{x}-1=0$ or $2\cos\text{x}+3=0$ $\Rightarrow\cos\text{x}=\frac{1}{2}$or $\cos\text{x}=-\frac{3}{2}$ [This is not possible as $-1<\cos\text{x}<1$] $\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$ $\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$
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