TI-(x 2x 5x) {solved}

[Speed2001]

 

[Lokman Gökçe]

x = 6^o

Solution:
m(BDC) = 3x, m(BCD) = 5x. So, 3x + 5x < 180^o and x < 22,5^o.

Now we apply sine theorem triangles ABD and BCD:

sin(180 - 3x)/sin2x = AB/BD, sin(180 - 8x)/sin5x = DC/BD.

Therefore: sin2x.sin8x = sin3x.sin5x. We know that graph of y = sin2x.sin8x - sin3x.sin5x and x axis intersect only one point, for interval 0^o < x < 22,5^o.

That is equation sin2x.sin8x = sin3x.sin5x is only one solution for 0^o < x < 22,5^o. Easily we check that sin12.sin48 = sin18.sin30. Finally x = 6^o is unique solution of equation sin2x.sin8x = sin3x.sin5x.

[Speed2001]

Thx for you solution scarface , but the equation sin2x.sin8x = sin3x.sin5x is impossible to solve by algebra (or trigonometry and calculus) if 0<X<22,5. 

Saludos.