Solving R-form Equations | CAIE 9709 Pure Mathematics 3 Notes

Pure Mathematics 3

Mathematics9709

Solve for the shifted angle before undoing the shift

After combining the two terms, divide by $R$ . Check whether the required sine or cosine lies between −1 and 1. Then transform the whole interval and find every angle in it.

Worked example

Solve 5 sin θ − 12 cos θ = 6 for 0° ≤ θ ≤ 360°.

\begin{gathered}5\sin\theta-12\cos\theta\\=13\sin(\theta-\alpha)\\\alpha=\tan^{-1}(12/5)\\u=\theta-\alpha\\-\alpha\le u\le360^\circ-\alpha\\\sin u=\frac6{13}\end{gathered}

Let $\beta=\sin^{-1}(6/13)$ . In the transformed interval, the two sine solutions are $u=\beta$ and $u=180^\circ-\beta$ . Add $\alpha$ to return to $\theta$ .

\begin{gathered}\theta=\alpha+\beta\\\text{or}\quad\theta=\alpha+180^\circ-\beta\end{gathered}

Answer

\theta\approx94.9^\circ,\ 219.9^\circ

These are to one decimal place, using full stored values of both angles. Substitution into the original expression gives 6.

Your turnPractice

Solve $\sqrt3\sin x+\cos x=1$ for $0^\circ\le x\le360^\circ$ . Give exact answers, including endpoints.

Show worked answer

\begin{gathered}2\sin(x+30^\circ)=1\\u=x+30^\circ\\30^\circ\le u\le390^\circ\\u=30^\circ,\ 150^\circ,\ 390^\circ\end{gathered}

Answer

x=0^\circ,\ 120^\circ,\ 360^\circ

The repeated sine value occurs at both included endpoints. List both because they are different values of $x$ .

Keep radians and scaled inputs throughout

For a radian interval, find the phase in radians too. If the input is $2x$ , the combined angle is $2x\pm\alpha$ ; undo both operations after solving.

Worked example

Solve √3 cos(2x) + sin(2x) = 1 for 0 ≤ x ≤ π.

\begin{gathered}2\cos(2x-\pi/6)=1\\u=2x-\frac\pi6\\-\frac\pi6\le u\le\frac{11\pi}6\\\cos u=\frac12\\u=\frac\pi3,\frac{5\pi}3\\x=\frac{u+\pi/6}{2}\end{gathered}

Answer

x=\frac\pi4,\ \frac{11\pi}{12}

Your turnPractice

Solve $\sin(x/2)+\sqrt3\cos(x/2)=2$ for $0\le x\le2\pi$ .

Show worked answer

\begin{gathered}2\sin(x/2+\pi/3)=2\\ \frac\pi3\le u=x/2+\frac\pi3\le\frac{4\pi}3\\ u=\frac\pi2\end{gathered}

Answer

x=\frac\pi3

At sine 1, the two usual sine branches coincide; there is only one distinct solution here.

Solve for the shifted angle before undoing the shift

After combining the two terms, divide by $R$ . Check whether the required sine or cosine lies between −1 and 1. Then transform the whole interval and find every angle in it.

Worked example

Solve 5 sin θ − 12 cos θ = 6 for 0° ≤ θ ≤ 360°.

\begin{gathered}5\sin\theta-12\cos\theta\\=13\sin(\theta-\alpha)\\\alpha=\tan^{-1}(12/5)\\u=\theta-\alpha\\-\alpha\le u\le360^\circ-\alpha\\\sin u=\frac6{13}\end{gathered}

Let $\beta=\sin^{-1}(6/13)$ . In the transformed interval, the two sine solutions are $u=\beta$ and $u=180^\circ-\beta$ . Add $\alpha$ to return to $\theta$ .

\begin{gathered}\theta=\alpha+\beta\\\text{or}\quad\theta=\alpha+180^\circ-\beta\end{gathered}

Answer

\theta\approx94.9^\circ,\ 219.9^\circ

These are to one decimal place, using full stored values of both angles. Substitution into the original expression gives 6.

Your turnPractice

Solve $\sqrt3\sin x+\cos x=1$ for $0^\circ\le x\le360^\circ$ . Give exact answers, including endpoints.

Show worked answer

\begin{gathered}2\sin(x+30^\circ)=1\\u=x+30^\circ\\30^\circ\le u\le390^\circ\\u=30^\circ,\ 150^\circ,\ 390^\circ\end{gathered}

Answer

x=0^\circ,\ 120^\circ,\ 360^\circ

The repeated sine value occurs at both included endpoints. List both because they are different values of $x$ .

Keep radians and scaled inputs throughout

For a radian interval, find the phase in radians too. If the input is $2x$ , the combined angle is $2x\pm\alpha$ ; undo both operations after solving.

Worked example

Solve √3 cos(2x) + sin(2x) = 1 for 0 ≤ x ≤ π.

\begin{gathered}2\cos(2x-\pi/6)=1\\u=2x-\frac\pi6\\-\frac\pi6\le u\le\frac{11\pi}6\\\cos u=\frac12\\u=\frac\pi3,\frac{5\pi}3\\x=\frac{u+\pi/6}{2}\end{gathered}

Answer

x=\frac\pi4,\ \frac{11\pi}{12}

Your turnPractice

Solve $\sin(x/2)+\sqrt3\cos(x/2)=2$ for $0\le x\le2\pi$ .

Show worked answer

\begin{gathered}2\sin(x/2+\pi/3)=2\\ \frac\pi3\le u=x/2+\frac\pi3\le\frac{4\pi}3\\ u=\frac\pi2\end{gathered}

Answer

x=\frac\pi3

At sine 1, the two usual sine branches coincide; there is only one distinct solution here.