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

8 0

2 years ago

How do I find x? Do I find the hypotenuse of 15 and 18, then with that answer minus it by 22^2 in a root?

pogonyaev

8+x= $\sqrt{22^2-15^2}$ => 8+x = 15 => x = 15-8=7

7 0

3 years ago