Solving $3\sin(2x+45^\circ)=2\cos(x+135^\circ)$ for $x$ between $0^\circ$ and $360^\circ$ Please find the value of $x$ in degree from this equation, with explanation
$$3\sin(2x+45^\circ)=2\cos(x+135^\circ)$$
For $x$ between $0^\circ$ and $360^\circ$.
 A: Your simplification of
$$3\sin(2x+45^\circ)=2\cos(x+135^\circ)$$
obtaining the left hand side of
$$3\sin2x\cos 45^\circ+3\cos 2x\sin 45^\circ =2\cos x\cos135^\circ-2\sin x\sin135^\circ$$
was very well done.
In fact, $\cos135^\circ=-\cos45^\circ$ and $\sin135^\circ=\sin45^\circ$ and so you can cancel down to
$$3\sin2x+3\cos 2x=-2\cos x-2\sin x$$
Using the double angle formulae you mention gives
$$6\sin x\cos x+6\cos^2 x-3=-2\cos x-2\sin x.$$
Check that you can follow these steps and that should suffice for your tutor until more work has been done on this topic.
A: Starting from S. Dolan's answer
$$6\sin x\cos x+6\cos^2 x-3=-2\cos x-2\sin x$$ let $c=\cos x$ ans $s=\sin x$ to make
$$6sc+6c^2-3=-2c-2s$$
Solving for $s$ gives $$s=\frac{-6 c^2-2 c+3}{2 (3 c+1)}$$, Square both sides and remember that $c^2+s^2=1$ to get
$$1-c^2=\left(\frac{-6 c^2-2 c+3}{2 (3 c+1)} \right)^2$$ Assuming $3c+1 \neq 0$, cross multiply, expand and simplify to get
$$72 c^4+48 c^3-64 c^2-36 c+5=0$$ which could be solved with (nasty) radicals (have a look here).
I wonder if there could be a typo in the problem since the solutions are quite ugly.
