Question

if 3sin2x=2sin3x and0<x<π , then value of sin2x is

Answer 1

Final Answer : sin 2x = 0,-√15/8

Understanding Formula :
1) sin 3x = sin x (2cos 2x + 1 )
2) sin 2x = 2sin x cos x
3) cos 2x = 2cos^2 (x )- 1

Steps :
1) 3 sin 2x = 2 sin 3x
$3 \sin(2x) = 2 \sin(3x) \\ = > 3 \times 2 \sin(x) \cos(x) = 2 \times \sin(x) \\ \times (2 \cos(2x) + 1) \\ = > \sin(x) = 0 \: \: - - (1) \: \: or \: \\ = > 3 \cos(x) = (2 \times 2 { (\cos(x) ) }^{2} - 2 + 1) \\ = > \: let \: \cos(x) = t \: \: \: \: then \: \\ 4 {t}^{2} - 3t - 1 = 0 \\ = > (4t + 1)(t - 1) = 0 \\ = > t = 1 \: \: or \: \: t \: = \frac{ - 1}{4}$
3) Equations we get:
$\sin(x) = 0$
$\cos(x) = 1 \: = > x = 0$

When, sin x = 0 ,
then sin 2x = 2 sin x cos x = 2 * 0 * 1 =0

When cos x =
$\frac{ - 1}{4} \: \: then \: \\ \sin(x) = + \sqrt{1 - {( \frac{ - 1}{4} }^{2} )} = \frac{ \sqrt{15} }{4} \\ \: \: or \: \frac{ - \sqrt{ 15} }{4}$

But sin x is positive in 0 to π ,
4) Then., sin 2 x = 2 sin x cos x
= 2 * √15 /4 * -1/4
= -√15/8

Therefore,
finally sin 2x can have values 0 , -√15/8