Question-86946

Question Number 86946 by A8;15: last updated on 01/Apr/20

Commented by john santu last updated on 01/Apr/20

A^{2} \sin^{2} Ax = B^{2} \sin Bx

\Rightarrow \sin^{2} Ax = \frac{B^{2}}{A^{2}} \sin^{2} Bx

\cos^{2} Ax = \cos^{2} Bx

\Rightarrow 1 - \sin^{2} Ax = 1 - \sin^{2} Bx

\Rightarrow \frac{B^{2}}{A^{2}} \sin^{2} Bx = \sin^{2} Bx

\Rightarrow B = A \lor B = - A

Commented by john santu last updated on 01/Apr/20

A = B \Rightarrow 2 \cos Ax = 0

\cos Ax = 0 \Rightarrow Ax = \pm \frac{π}{2} + 2 k π

x = \pm \frac{π}{2 A} + \frac{2 k π}{A}