Question

2- cos x = 2 tan x/2 general solution​

Answer 1

Step-by-step explanation:

We have,

$2 - \cos(x) = 2 \tan \bigg( \frac{x}{2} \bigg)$

$\implies2 - \frac{1 - \tan^{2} \bigg( \dfrac{x}{2} \bigg) }{1 + \tan^{2} \bigg( \dfrac{x}{2} \bigg) } = 2 \tan \bigg( \frac{x}{2} \bigg)$

$\implies \frac{2 \bigg \{ 1 + \tan^{2} \bigg( \dfrac{x}{2} \bigg) \bigg \} - 1 + \tan^{2} \bigg( \dfrac{x}{2} \bigg) }{1 + \tan^{2} \bigg( \dfrac{x}{2} \bigg) } = 2 \tan \bigg( \frac{x}{2} \bigg)$

$\implies \frac{2 + 2\tan^{2} \bigg( \dfrac{x}{2} \bigg) - 1 + \tan^{2} \bigg( \dfrac{x}{2} \bigg) }{1 + \tan^{2} \bigg( \dfrac{x}{2} \bigg) } = 2 \tan \bigg( \frac{x}{2} \bigg)$

$\implies \frac{1 + 3\tan^{2} \bigg( \dfrac{x}{2} \bigg) }{1 + \tan^{2} \bigg( \dfrac{x}{2} \bigg) } = 2 \tan \bigg( \frac{x}{2} \bigg)$

$\implies 1 + 3\tan^{2} \bigg( \dfrac{x}{2} \bigg) = 2 \tan \bigg( \frac{x}{2} \bigg) \bigg \{ 1 + \tan^{2} \bigg( \dfrac{x}{2} \bigg) \bigg \}$

$\implies 1 + 3\tan^{2} \bigg( \dfrac{x}{2} \bigg) = 2 \tan \bigg( \frac{x}{2} \bigg) + 2\tan^{3} \bigg( \dfrac{x}{2} \bigg)$

$\implies 2\tan^{3} \bigg( \dfrac{x}{2} \bigg) - 3\tan^{2} \bigg( \frac{x}{2} \bigg) + 2 \tan \bigg( \frac{x}{2} \bigg) - 1 = 0$

$\implies 2\tan^{3} \bigg( \dfrac{x}{2} \bigg) - 2\tan^{2} \bigg( \frac{x}{2} \bigg) - \tan^{2} \bigg( \frac{x}{2} \bigg)+ \tan \bigg( \frac{x}{2} \bigg) + \tan \bigg( \frac{x}{2} \bigg) - 1 = 0$

$\implies 2\tan^{2} \bigg( \frac{x}{2} \bigg) \bigg \{ \tan \bigg( \dfrac{x}{2} \bigg) - 1 \bigg \} - \tan \bigg( \frac{x}{2} \bigg) \bigg \{ \tan \bigg( \frac{x}{2} \bigg) - 1 \bigg \} + 1 \bigg \{ \tan \bigg( \frac{x}{2} \bigg) - 1 \bigg \} = 0$

$\implies \bigg \{ \tan \bigg( \frac{x}{2} \bigg) - 1 \bigg \} \bigg \{ 2\tan^{2} \bigg( \frac{x}{2} \bigg) - \tan \bigg( \frac{x}{2} \bigg) + 1 \bigg \} = 0$

$\implies \bigg \{ \tan \bigg( \frac{x}{2} \bigg) - 1 \bigg \} \bigg \{ 2\tan^{2} \bigg( \frac{x}{2} \bigg) - 2\tan \bigg( \frac{x}{2} \bigg) + \tan \bigg( \frac{x}{2} \bigg) + 1 \bigg \} = 0$

$\implies \bigg \{ \tan \bigg( \frac{x}{2} \bigg) - 1 \bigg \} \bigg [ 2\tan \bigg( \frac{x}{2} \bigg) \bigg \{ \tan \bigg( \frac{x}{2} \bigg) - 1 \bigg \} +1 \bigg \{ \tan \bigg( \frac{x}{2} \bigg) + 1 \bigg \} \bigg] = 0$

$\implies \bigg \{ \tan \bigg( \frac{x}{2} \bigg) - 1 \bigg \} \bigg \{ 2\tan \bigg( \frac{x}{2} \bigg) + 1 \bigg \} \bigg \{ \tan \bigg( \frac{x}{2} \bigg) - 1 \bigg \} = 0$

$\implies \bigg \{ \tan \bigg( \frac{x}{2} \bigg) - 1 \bigg \}^{2}. \bigg \{ 2\tan \bigg( \frac{x}{2} \bigg) + 1 \bigg \} = 0$

$\rm \: Either \: \: \tan \bigg( \frac{x}{2} \bigg) - 1 = 0 \: \: \: or \: \: \: 2\tan \bigg( \frac{x}{2} \bigg) + 1 = 0$

$\rm \implies \: \: \tan \bigg( \frac{x}{2} \bigg) = 1 \: \: \: or \: \: \: 2\tan \bigg( \frac{x}{2} \bigg) = - 1$

$\rm \implies \: \: \tan \bigg( \frac{x}{2} \bigg) = 1 \: \: \: or \: \: \: \tan \bigg( \frac{x}{2} \bigg) = - \frac{ 1 }{2}$

$\rm \implies \: \: \frac{x}{2} = n\pi + \frac{\pi}{4} \: \: \: or \: \: \: \frac{x}{2} = m\pi - \tan^{ - 1} \bigg( \frac{ 1 }{2} \bigg)$

$\rm \implies \: \: x = 2n\pi + \frac{\pi}{2} \: \: \: or \: \: \: x = 2m\pi - 2\tan^{ - 1} \bigg( \frac{ 1 }{2} \bigg)$

$\rm \implies \: \: x = (4n + 1) \frac{\pi}{2} \: \: \: or \: \: \: x = 2m\pi - \tan^{ - 1} \bigg \{ \frac{ 2 \times \frac{1}{2} }{1 - {( \frac{1}{2}) }^{2} } \bigg \}$

$\rm \implies \: \: x = (4n + 1) \frac{\pi}{2} \: \: \: or \: \: \: x = 2m\pi - \tan^{ - 1} \bigg \{ \frac{ 1 }{1 - \frac{1}{4} } \bigg \}$

$\rm \implies \: \: x = (4n + 1) \frac{\pi}{2} \: \: \: or \: \: \: x = 2m\pi - \tan^{ - 1} \bigg \{ \frac{ 1 }{ \frac{4 - 1}{4} } \bigg \}$

$\rm \implies \: \: x = (4n + 1) \frac{\pi}{2} \: \: \: or \: \: \: x = 2m\pi - \tan^{ - 1} \bigg \{ \frac{ 1 }{ \frac{3}{4} } \bigg \}$

$\rm \implies \: \: x = (4n + 1) \frac{\pi}{2} \: \: \: or \: \: \: x = 2m\pi - \tan^{ - 1} \bigg ( \frac{ 4 }{ 3 } \bigg)$