Question

Sin 3x = 1 și x aparine (0, pi/6 )

calculati: cos 2 x + cos 4 x ​

Answer 1

$\sin3x = 1 \: \: \: \: \: \: = > 3x = \frac{\pi}{2} \\ x \: e(0. \frac{\pi}{6} ) \\ \\ \cos(2x) = { \cos}^{2} x - { \ \sin}^{2} x \\ \sin \frac{\pi}6 = \frac{1}{2} \\ \cos \frac{\pi}6 = \frac{ \sqrt{3} }{2} \\ \cos \: 2x = \frac{3}{4} - \frac{1}{4} = \frac{2}{4} = \frac{1}{2} \\ { \cos}^{2} x \: = \frac{1}{4} \\ \cos4x = 2 \cos ^{2} 2x - 1 = 2 \times \frac{1}{4} - 1 = \frac{2}{4} - 1 = \frac{1}{2} - 1 = \frac{1 - 2}{2} = - \frac{1}{2} \\ \cos2x \ + \cos4x = \frac{1}{2} + ( - \frac{ 1}{2} ) = 0 \: \\ apartine \: (0. \frac{\pi}{6} )$