<s>[INST] 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 $360^\circ$.
 [/INST] 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.
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