Question

9sin(2x) sin (x) = 9cos(x)

Answer 1

`\sin(2x) = 2 \sin(x) \cos(x) \\ 9( 2 \sin(x) \cos(x) ) \sin(x) = 9 \cos(x) \\ 18 { \sin}^(2) (x)9 \cos(x) - 9 \cos(x) = 0 \\ 9 \cos(x) (2{ \sin}^(2) (x) - 1) = 0 \\ 9 \cos(x) = 0 \: or \: 2{ \sin}^(2) (x) - 1 = 0 \\ \sin(x) = \pm (1)/( √(2) ) \\ \cos(x) = 0 \\ x = { \cos}^( - 1) (0) \\ x = ( \pi)/(2) = 90 \degree \\ x = ( \pi)/(2) + 2\pi \: n \: \forall \: n \: \in \: \Z \\ = 90 \degree + 360\degree\: n \: \forall \: n \: \in \: \Z \\ = \pm( \pi)/(4) \pm2\pi \: n \: \forall \: n \: \in \: \Z \\ x = \pm45\degree \pm 360\degree\: n \: \forall \: n \: \in \: \Z`