Question

Solve 2sin^2 (x + π/3) = 1 for 0<x<2π radians​

Answer 1

Step-by-step explanation:

$2 {sin}^{2} (x + \frac{\pi}{3}) = 1 \\ 1 - 2 {sin}^{2} (x + \frac{\pi}{3} ) = 0 \\ cos2( x + \frac{ \pi }{3} ) = 0 \\ cos(2x + \frac{2\pi}{3} ) = 0 \\ \cos(2x) \cos( \frac{2\pi}{3} ) - \sin(2x) \sin( \frac{2\pi}{3} ) = 0 \\ cos(2x) \cos( \frac{2\pi}{3} ) = \sin(2x) \sin( \frac{2\pi}{3} ) \\ \cos(2x) \cos(\pi - \frac{\pi}{3} ) = \sin(2x) \sin(\pi - \frac{\pi}{3} ) \\ \cos(2x) ( - \cos( \frac{\pi}{3} ) ) = \sin(2x) \sin( \frac{\pi}{3} ) \\ \cos(2x) ( - \frac{1}{2} ) = \sin(2x) ( \frac{ \sqrt{3} }{2}) \\ \frac{ \sin(2x) }{ \cos(2x) } = \frac{ \frac{ - 1}{2} }{ \frac{ \sqrt{3} }{2} } \\ \tan(2x) = - \frac{1}{ \sqrt{3 } } \\ \\ now \: \tan( \frac{\pi}{6} ) = \frac{1}{ \sqrt{3} } \\ \tan(2 \times \frac{\pi}{12} ) = \frac{1}{ \sqrt{3} } \\ \tan( \frac{10\pi}{12} ) = - \frac{1}{ \sqrt{3} }$

x = 10π/12