Question

solve secx-tanx=✓3 hint (cosx does not equal to 0)​

Answer 1

Answer:

$sec \: x - tan \: x = \sqrt{3} \\ \therefore \: \frac{1}{cos \: x} - \frac{sin \: x}{cos \: x} = \sqrt{3} \\ \therefore \: 1 - sin \: x = \sqrt{3} cos \: x \\squaring \: both \: sides \\ (1 - sin \: x)^{2} = ( \sqrt{3} \: cos \: x)^{2} \\ \therefore \: 1 + {sin}^{2} \: x - 2 \: sin \: x = 3 \: cos^{2} \: x \\ \therefore \: 1 + {sin}^{2} \: x - 2 \: sin \: x = 3(1 - sin^{2} \: x) \\ \therefore \: 1 + {sin}^{2} \: x - 2 \: sin \: x = 3 - 3sin^{2} \: x \\ \therefore \: 1 + {sin}^{2} \: x - 2 \: sin \: x - 3 + 3sin^{2} \: x = 0 \\ \therefore \: \ 4sin^{2} \: x - 2 \: sin \: x - 2 = 0 \\ \therefore \: \ 4sin^{2} \: x - 4\: sin \: x + 2 \: sin \: x- 2 = 0 \\ \therefore \:4 \: sin \: x(sin \: x - 1) + 2(sin \: x - 1) = 0 \\ \therefore \:(sin \: x - 1)(4 \: sin \: x+ 2) = 0 \\ \therefore \:(sin \: x - 1) = 0 \: \: or \: \: (4 \: sin \: x+ 2) = 0 \\ \therefore \:sin \: x = 1 \: \: or \: \: \:sin \: x = - \frac{2}{4} \\ \therefore \:sin \: x = 1 \: \: or \: \: \:sin \: x = - \frac{1}{2} \\ \therefore \:sin \: x = sin \: \frac{ \pi}{2} \: \: or \: \: \:sin \: x = sin (\pi + \frac{\pi}{6} ) \\ \therefore \:x = \frac{ \pi}{2} \: \: or \: \: x = \frac{7\pi}{6}$