Originally Posted by

**flyingsquirrel** Hi,

I think it is correct too.

I'd rather do it this way :

2\sin x -3\cos x &= \sqrt{2^2+3^2}\left(\frac{2}{\sqrt{2^2+3^2}}\sin x -\frac{3}{\sqrt{2^2+3^2}}\cos x\right)\\

&=\sqrt{13}\left(\frac{2}{\sqrt{13}}\sin x -\frac{3}{\sqrt{13}}\cos x\right)

\end{aligned}

" alt="

2\sin x -3\cos x &= \sqrt{2^2+3^2}\left(\frac{2}{\sqrt{2^2+3^2}}\sin x -\frac{3}{\sqrt{2^2+3^2}}\cos x\right)\\

&=\sqrt{13}\left(\frac{2}{\sqrt{13}}\sin x -\frac{3}{\sqrt{13}}\cos x\right)

\end{aligned}

" />

This is interesting because

so there exists

such that

. This gives us

2\sin x -3\cos x = \sqrt{13}\left(\sin x\cos \phi -\sin\phi\cos x\right)

=\sqrt{13}\sin(x-\phi)

" alt="

2\sin x -3\cos x = \sqrt{13}\left(\sin x\cos \phi -\sin\phi\cos x\right)

=\sqrt{13}\sin(x-\phi)

" />

Can you take it from here ?

Yes, this equation has no solution.

Thanks but I don't think my lil cousin will understand that method when I try to explain it to him. I approached it like this

square both sides

Then I go ahead and use u of substitation for the cosx and get

and I get 2.43 and 1.26. Then when I check it graphically it tells me my only solution is 1.26 and 3.84. Where am I going wrong and once again thanks in advance.