Question

sin2xcosx + cos2xsinx=-1/2

Answer 1

remember
\(\sin(A+B) = \sin A\cos B+\cos A \sin B \)

Answer 2

I need to find the solutions in the interval (0,2pi) for the equation

Answer 3

before finding the solution, did u use the given identity and simplified which leads to the solution?

Answer 4

In order to solve this, we have to know some trigonometric identities:
\[\cos2x=\cos^2x-\sin^2x=1-\sin^2x=2\cos^2x-1\]
\[\sin2x=2senxcosx\]
and use them to take all the double angles, and convert the whole equation into a single trigonometric function.
or we could use, as Bahdi said, the sum of of angles.
\[Sin(A+B)=sinAcosB+cosAsinB\]
And, we can see, that the structure actually works, so, given:
\[\sin2xcosx+\cos2xsinx=-\frac{ 1 }{ 2 }\]
if we call A=2x and B=x, then:
\[sinAcosB+cosAsinB=-\frac{ 1 }{ 2 }\]
and, transforming the identity, we can see that:
\[\sin(A+B)=-\frac{ 1 }{ 2 }\]
but, we said that A=2x, and B=x
So therefore:
\[\sin(2x+x)=-\frac{ 1 }{ 2 }\]
I'll let you continue from this point.