Question

Find the General Solution of \[\sin^2 x + \sin^2 2x = \sin^2 3x\]?

Answer 1

@shubhamsrg @satellite73 @RadEn

Answer 2

apply there identities, you'll get

sin^2 x + 4 sin^2 x cos^2 x = (-4sin^3 x + 3sinx)^2 = sin^2 x( 3 - 4sin^2 x)^2

one clear solution will be
sin^2 x=0 or sin x =0 ..generalize this

we are left with

1 + 4 cos^2 x = 9 + 16 sin^4 x - 24 sin^2 x
=>1 + 4 - 4sin^2 x = 9 + 16 sin^4 x - 24 sin^2x
=> 16sin^4 x - 20 sin^2 x + 4 =0
=> 4 sin^4 x - 5 sin^2 x + 4 =0

its a quadratic eqn in sin^2 x which can be easily solved (5= 4+1)

rest should be easier..

Answer 3

sin^2 x = sin^2 x :)
sin^ 2x = (2sinxcosx)^2 = 4sin^2 x cos^2 x = 4sin^2 x - 4sin^4 x
(sin3x)^2 = (-4sin^3 x + 3sinx)^2 = 16sin^6 x - 24sin^4 x + 9sin^2 x
now, u get the equation in sinx form... rearreange giving us :
16sin^6 x -20sin^4 x + 4sin^2 x = 0 or
4sin^6 x -5sin^4 x + sin^2 x = 0
sin^2 x (4sin^4 x - 5sin^2 x + 1) = 0
sin^2 x (4sin^2 x - 1)(sin^2 x - 1) = 0
sin^2 x (2sinx+1)(2sinx-1)(sinx+1)(sinx-1) = 0
now, it looks be beatifull form right ?