Question

Find number of solution of sin^2 x + sin^2 2x = sin^2 3x , if -90 <= x <= 180

Answer 1

it is given that,

sin²x + sin²2x = sin²3x

or, sin²2x = sin²3x - sin²x

or, sin²2x = (sin3x - sinx)(sin3x + sinx)

we know, sinC - sinD = 2cos(C + D)/2.sin(C - D)/2

sinC + sinD = 2sin(C + D)/2.cos(C - D)/2

or, sin²2x = (2cos2x.sinx)(2sin2x.cosx)

or, sin²2x = (2sin2x.cos2x)(2sinx.cosx)

or, sin²2x = (sin4x)(sin2x)

or, sin2x(sin2x - sin4x) = 0

or, sin2x(2cos3x.sin(-x)) = 0

or, sinx.sin2x.cos3x = 0

we get, sinx = 0 ⇒x = 0°, 180°

sin2x = 0 ⇒x = -90°, 0°, 90°, 180°

cos3x = 0 ⇒3x = -90°, 90°, 270°, 450° , x = -30° , 30° , 90°, 150°

so, x = -90°, -30°, 0°, 30°, 90°, 150°, 180°

hence, total number of solutions = 7