how do I solve this equation ?

Question

amoodarya · Answer

$$3\sin^3x+2\cos^3x=2sinx+cosx$$

mathslover · Answer

I'll be more than fascinated to see what you've tried yet, amoodarya.

amoodarya · Answer

I need to solve it analytical ! not by graph ,
I know one of roots is $$\frac{\pi}{3}$$

mathslover · Answer

I wonder, were you given with that root or did you calculate it on your own?

amoodarya · Answer

I test it , but I can't solve it nice !

mathslover · Answer

One method is to follow the conventional way - that is putting $\sin x = \sqrt{1- \cos^2 x} $ 

Although, this is complicated. But, are you studying topic related to systems of equations (or similar to this) ...?

mathslover · Answer

And if one of the roots is $\pi/3 $ then : 

$3(3\sqrt{3} /8 ) + 2 (1/8) = 2(\qrt{3})/2 + 1/2 \
9\sqrt{3} /8 + 1/4  = \sqrt{3} + 1/2 $ 

This root doesn't seem to satisfy the given equation.

anonymous · Answer

At first sight, I can't see a nice solution to this problem, maybe Weierstrass substitution will work

amoodarya · Answer

i solve it numerically , then I saw one root is very close to pi/3 , but missed to check it in equation .
thanks for your notification "
mukushla"

anonymous · Answer

no problem amoodarya

ganeshie8 · Answer

is this you @amoodarya http://math.stackexchange.com/questions/1344917/how-do-i-solve-this-equation-mathbbr-3-sin3x2-cos3x-2-sin-x-cos