Solve the following equation: 4 ^ 2 x-8 +1=0 - Vidyadip

Question

Solve the following equation:
$4\sin^{2}\text{x}-8\cos\text{x}+1=0$

✓

Answer

We have,

$4\sin^{2}\text{x}-8\cos\text{x}+1=0$

$\Rightarrow4(1-\cos^{2}\text{x})-8\cos\text{x}+1=0$

$\Rightarrow4\cos^{2}\text{x}+8\cos\text{x}-5=0$

factorise it,we get,

$\Rightarrow4\cos^{2}\text{x}+10\cos\text{x}-2\cos\text{x}-5=0$

$\Rightarrow2\cos\text{x}(2\cos\text{x}+5)-1(2\cos\text{x}+5)=0$

$\Rightarrow(2\cos\text{x}-1)(2\cos\text{x}+5)=0$

Either $2\cos\text{x}-1=0$ or $2\cos\text{x}+5=0$

$\Rightarrow\cos\text{x}=\frac{1}{2}$ or $\cos\text{x}=-\frac{5}{2}$

$\Rightarrow\cos\text{x}=\cos\frac{\pi}{3}$

$\Rightarrow\text{x}=2\text{n}\pi\pm\frac{\pi}{3},\text{n}\in\text{z}$

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