Find the values of \(x\) between \(0^\circ\) and \(360^\circ\) which satisfy the equation

\[3\cos x + 1 = 2\sin x.\]

We can rearrange our equation to \[2\sin x- 3\cos x = 1.\]

Now we can use the compound angle identity to rewrite the left hand side as \[2\sin x- 3\cos x = R\sin(x-\alpha) = R\sin x\cos\alpha - R\cos x\sin\alpha\] and matching the coefficients tells us we need \(R\) and \(\alpha\) such that \[R\cos\alpha=2 \quad\text{and}\quad R\sin\alpha=3.\] This reduces to \[\begin{align*} R^2=13 \quad&\text{and}\quad \tan\alpha=\frac{3}{2} \\ \implies\quad R=\sqrt{13} \quad&\text{and}\quad \alpha=56.3^\circ (1 \text{d.p.}) \end{align*}\]

So now our original equation becomes \[\sin(x-56.3^\circ)=\frac{1}{\sqrt{13}}.\]

The general solution to this is \[x=\arcsin\frac{1}{\sqrt{13}}+56.3^\circ \quad\text{or}\quad x=180^\circ - \arcsin\frac{1}{\sqrt{13}}+56.3^\circ\] plus an integer multiple of \(360^\circ\).

Within the required range this gives us \[x=72.4^\circ\quad\text{or}\quad 220.2^\circ (1 \text{d.p.}).\]

Check – if we have access to a graphing program, we can see where the graphs of \(y = 3\cos x + 1\) and \(y = 2\sin x\) intersect.

the graphs of y = 3cos x + 1 and y = 2sin x