Review question

# Can we solve the equation $3\cos x + 1 = 2\sin x$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5113

## Solution

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.