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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   ⇒sin^2  Ax = (B^2 /A^2 ) sin^2 Bx  cos^2 Ax = cos^2 Bx   ⇒1−sin^2 Ax = 1−sin^2 Bx  ⇒(B^2 /A^2 ) sin^2 Bx = sin^2 Bx  ⇒B = A ∨ B = −A
$$\mathrm{A}^{\mathrm{2}} \:\mathrm{sin}\:^{\mathrm{2}} \mathrm{Ax}\:\:=\:\mathrm{B}^{\mathrm{2}} \:\mathrm{sin}\:\mathrm{Bx}\: \\ $$$$\Rightarrow\mathrm{sin}\:^{\mathrm{2}} \:\mathrm{Ax}\:=\:\frac{\mathrm{B}^{\mathrm{2}} }{\mathrm{A}^{\mathrm{2}} }\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{Bx} \\ $$$$\mathrm{cos}\:^{\mathrm{2}} \mathrm{Ax}\:=\:\mathrm{cos}\:^{\mathrm{2}} \mathrm{Bx}\: \\ $$$$\Rightarrow\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{Ax}\:=\:\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{Bx} \\ $$$$\Rightarrow\frac{\mathrm{B}^{\mathrm{2}} }{\mathrm{A}^{\mathrm{2}} }\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{Bx}\:=\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{Bx} \\ $$$$\Rightarrow\mathrm{B}\:=\:\mathrm{A}\:\vee\:\mathrm{B}\:=\:−\mathrm{A} \\ $$$$ \\ $$
Commented by john santu last updated on 01/Apr/20
A = B ⇒ 2cos Ax = 0  cos Ax = 0 ⇒ Ax = ± (π/2) + 2kπ   x = ± (π/(2A)) + ((2kπ)/A)
$$\mathrm{A}\:=\:\mathrm{B}\:\Rightarrow\:\mathrm{2cos}\:\mathrm{Ax}\:=\:\mathrm{0} \\ $$$$\mathrm{cos}\:\mathrm{Ax}\:=\:\mathrm{0}\:\Rightarrow\:\mathrm{Ax}\:=\:\pm\:\frac{\pi}{\mathrm{2}}\:+\:\mathrm{2k}\pi\: \\ $$$$\mathrm{x}\:=\:\pm\:\frac{\pi}{\mathrm{2A}}\:+\:\frac{\mathrm{2k}\pi}{\mathrm{A}} \\ $$

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