Equations with Double Angles | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Factor a common ratio instead of dividing by it

Replace a double angle by its single-angle expression. If the resulting terms have a common factor, bring them to one side and factorise. Dividing by that factor could remove valid roots.

Worked example

Solve sin 2x = sin x for 0° ≤ x ≤ 360°.

\begin{gathered}2\sin x\cos x-\sin x=0\\\sin x(2\cos x-1)=0\\\sin x=0\\\text{or}\quad\cos x=\frac12\end{gathered}

A product is zero when at least one factor is zero. The first branch gives $0^\circ,180^\circ,360^\circ$ ; the second gives $60^\circ,300^\circ$ . Dividing by sine would lose all three roots in the first branch.

Answer

\begin{gathered}x=0^\circ,60^\circ,180^\circ,\\300^\circ,360^\circ\end{gathered}

Blue: sin 2x. Green dashed: sin x. Red points are the valid zero roots that cancellation would lose.

Your turnPractice

Solve $2\sin2x=\cos x$ for $0^\circ\le x<360^\circ$ .

Show worked answer

\cos x(4\sin x-1)=0

Cosine zero gives $90^\circ,270^\circ$ . Sine $1/4$ gives approximately $14.5^\circ,165.5^\circ$ . All four satisfy the original equation.

Choose the double-angle form before solving

Worked example

Solve cos 2x + 3 sin x = 2 for 0 ≤ x ≤ 2π.

The other term is sine, so use $\cos2x=1-2\sin^2x$ .

\begin{gathered}u=\sin x\\1-2u^2+3u=2\\(2u-1)(u-1)=0\\\sin x=\frac12\quad\text{or}\quad1\end{gathered}

Answer

x=\frac\pi6,\ \frac\pi2,\ \frac{5\pi}6

Sine 1 gives only one angle in this interval. The two usual sine branches coincide at the maximum; do not list it twice.

Your turnPractice

Solve $3\cos2x+\cos x=2$ for $0^\circ\le x\le180^\circ$ .

Show worked answer

\begin{gathered}u=\cos x\\6u^2+u-5=0\\(6u-5)(u+1)=0\end{gathered}

Answer

x\approx33.6^\circ,\qquad x=180^\circ

Both cosine values are in $[-1,1]$ . Square brackets include the endpoints; cosine −1 gives the allowed endpoint $180^\circ$ .

Factor a common ratio instead of dividing by it

Replace a double angle by its single-angle expression. If the resulting terms have a common factor, bring them to one side and factorise. Dividing by that factor could remove valid roots.

Worked example

Solve sin 2x = sin x for 0° ≤ x ≤ 360°.

\begin{gathered}2\sin x\cos x-\sin x=0\\\sin x(2\cos x-1)=0\\\sin x=0\\\text{or}\quad\cos x=\frac12\end{gathered}

Answer

\begin{gathered}x=0^\circ,60^\circ,180^\circ,\\300^\circ,360^\circ\end{gathered}

Blue: sin 2x. Green dashed: sin x. Red points are the valid zero roots that cancellation would lose.

Your turnPractice

Solve $2\sin2x=\cos x$ for $0^\circ\le x<360^\circ$ .

Show worked answer

\cos x(4\sin x-1)=0

Cosine zero gives $90^\circ,270^\circ$ . Sine $1/4$ gives approximately $14.5^\circ,165.5^\circ$ . All four satisfy the original equation.

Choose the double-angle form before solving

Worked example

Solve cos 2x + 3 sin x = 2 for 0 ≤ x ≤ 2π.

The other term is sine, so use $\cos2x=1-2\sin^2x$ .

\begin{gathered}u=\sin x\\1-2u^2+3u=2\\(2u-1)(u-1)=0\\\sin x=\frac12\quad\text{or}\quad1\end{gathered}

Answer

x=\frac\pi6,\ \frac\pi2,\ \frac{5\pi}6

Sine 1 gives only one angle in this interval. The two usual sine branches coincide at the maximum; do not list it twice.

Your turnPractice

Solve $3\cos2x+\cos x=2$ for $0^\circ\le x\le180^\circ$ .

Show worked answer

\begin{gathered}u=\cos x\\6u^2+u-5=0\\(6u-5)(u+1)=0\end{gathered}

Answer

x\approx33.6^\circ,\qquad x=180^\circ

Both cosine values are in $[-1,1]$ . Square brackets include the endpoints; cosine −1 gives the allowed endpoint $180^\circ$ .