Hi,

Originally Posted by

**11rdc11** Ok so for problem A I did this

(...) I believe this is correct

I think it is correct too.

Originally Posted by

**11rdc11** C) $\displaystyle 3\sin{x} - 5 = 0$

Originally Posted by

**11rdc11** For problem B I'm not sure where to start.

I'm guessing either square both sides or equal the right side to cos^2x + sin^2x =1

I'd rather do it this way :

$\displaystyle \begin{aligned}

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 $\displaystyle \left(\tfrac{2}{\sqrt{13}}\right)^2+\left(\tfrac{3 }{\sqrt{13}}\right)^2=1$ so there exists $\displaystyle \phi\in[0,2\pi[$ such that $\displaystyle \begin{cases}\sin \phi = \tfrac{3}{\sqrt{13}}\\ \cos \phi=\tfrac{2}{\sqrt{13}} \end{cases}$. This gives us

$\displaystyle

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 ?

Originally Posted by

**11rdc11** For problem C

I get that it is false due to range

Yes, this equation has no solution.