Question

$find \: general \: solution \: of \: the \: equation \: \sin^2 \alpha { - 2 \cos( \alpha ) } + \frac{1}{4} = 0$
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Answer 1

Step-by-step explanation:

We have,

$\sin^{2} ( \alpha ) - 2 \cos( \alpha ) + \frac{1}{4} = 0$

$\implies1 - \cos^{2} ( \alpha ) - 2 \cos( \alpha ) + \frac{1}{4} = 0$

$\implies4 - 4\cos^{2} ( \alpha ) - 8\cos( \alpha ) + 1 = 0$

$\implies5- 4\cos^{2} ( \alpha ) - 8\cos( \alpha ) = 0$

$\implies 4\cos^{2} ( \alpha ) + 8\cos( \alpha ) - 5= 0$

$\implies 4\cos^{2} ( \alpha ) + 10\cos( \alpha ) - 2 \cos( \alpha ) - 5= 0$

$\implies 2\cos ( \alpha ) (2 \cos( \alpha ) + 5) -1( 2 \cos( \alpha ) + 5)= 0$

$\implies (2\cos( \alpha ) - 1) (2 \cos( \alpha ) + 5) = 0$

$\implies (2\cos ( \alpha )- 1) = 0 \: \: or \: \: (2 \cos( \alpha ) + 5) = 0$

$\implies \cos ( \alpha ) = \frac{1}{2} \: \: or \: \: \cos( \alpha ) = - \frac{5}{2}$

Since $\cos(\alpha) \in [-1,1]$

so, $\cos(\alpha) ≠- \frac{5}{2}$

so,

$\cos( \alpha ) = \frac{1}{2}$

$\implies \: \cos( \alpha ) = \cos \bigg( \frac{\pi}{3} \bigg)$

$\implies \: \alpha = 2n\pi \pm \frac{\pi}{3}$