<s>[INST] Express $2\cos^2x - 3\sin^2x$ in terms of $\cos(2x)$ I need to express $2\cos^2x - 3\sin^2x$ in terms of $\cos(2x)$.
I have substituted the $\sin^2x$ using trig identities and have reached:
$5\cos^2x - 3$.
But now I'm really stuck.
 [/INST] So $2\cos^{2}x-3\sin^{2}x=5\cos^{2}x -3$ but also $\cos 2x=2\cos^{2}x-1$ substitution give \begin{align*} 2\cos^{2}x-3\sin^{2}x&=5\cos^{2}x -3,\\&=5\left(\frac{1}{2}+\frac{1}{2}\cos2x\right)-3,\\&=-\frac{1}{2}+\frac{5}{2}\cos 2x.\end{align*}
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