Question
Solve the following equation:
$\sin^{2}\text{x}-\cos\text{x}=\frac{1}{4}$
$\sin^{2}\text{x}-\cos\text{x}=\frac{1}{4}$
✓
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 $-1<\cos\text{x}<1$]
$\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$
$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$
$\sin^{2}\text{x}-\cos\text{x}=\frac{1}{4}$
$\Rightarrow1-\cos^{2}\text{x}-\cos\text{x}=\frac{1}{4}$ $[\because\sin^{2}\text{x}=1-\cos^{2}\text{x}]$
$\Rightarrow\cos^{2}\text{x}+\cos\text{x}-\frac{3}{4}=0$
$\Rightarrow4\cos^{2}\text{x}+4\cos\text{x}-3=0$
$\Rightarrow4\cos^{2}\text{x}+6\cos\text{x}+2\cos\text{x}-3=0$ [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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