Can we solve the equation $3\cos x + 1 = 2\sin x$?

Can we solve the equation $3\cos x + 1 = 2\sin x$?

Question

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

$$3\cos x + 1 = 2\sin x.$$

Suggestion

Could we rewrite $\sin (x-\alpha)$ in a helpful way?

How many solutions would you expect to find in the range?

Can you sketch a graph to illustrate the equation?

Solution

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.