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.