Quadratic Trig Equations | CAIE 9709 Pure Mathematics 1 Notes

Pure Mathematics 1

Mathematics9709

Solve for the ratio, then the angles

A quadratic can contain $\cos x$ instead of $x$ . Treat the entire cosine value as one unknown, then return to the angle.

Worked example

Solve $2\cos^2x-\cos x-1=0$ for $0^\circ\le x\le360^\circ$ .

Let $u=\cos x$ . Enter coefficients $2,-1,-1$ into the calculator's quadratic / degree-2 polynomial solver. The roots are $u=1$ and $u=-1/2$ , so write:

2u^2-u-1=0

(u-1)(2u+1)=0

The factor 2 is needed to keep the leading coefficient. Expanding checks the working. Now return to cosine:

\cos x=1\quad\text{or}\quad\cos x=-\frac12

The first gives 0° and 360°. The second gives 120° and 240°.

Answer

x=0^\circ,\ 120^\circ,\ 240^\circ,\ 360^\circ

Convenient roots let you write factorised working directly. If the calculator gives awkward roots, use the quadratic formula for exact working; do not guess factors from rounded decimals. If the question names a method, use that method.

Before pressing an inverse key, reject sine or cosine values outside $[-1,1]$ . Tangent has no such range restriction. Every valid ratio root needs its own angles.

Factor before dividing

For $2\sin^2x-\sin x=0$ , dividing by sine would lose every solution where sine is zero. Factor instead:

\sin x(2\sin x-1)=0

A product is zero when at least one factor is zero. Solve both $\sin x=0$ and $\sin x=1/2$ . On $0^\circ\le x\le360^\circ$ , that gives 0°, 30°, 150°, 180°, 360°.

Be careful with a squared substitution too. If $u=\sin^2x$ , then $0\le u\le1$ . A root $u=1/4$ gives $\sin x=\pm1/2$ , because both signs square to $1/4$ .

Your turn

Your turn9709/12 Oct/Nov 2021 Q1 [4 marks]

Solve for $-90^\circ<\theta<90^\circ$ :

2\cos\theta=7-\frac3{\cos\theta}

Show worked answer

Cosine cannot be zero in the original fraction. Multiply by cosine and bring all terms to one side:

2\cos^2\theta-7\cos\theta+3=0

The polynomial solver gives $1/2$ and 3. The supporting factorisation is:

(2\cos\theta-1)(\cos\theta-3)=0

Reject 3 because cosine cannot exceed 1. From $\cos\theta=1/2$ , use the two angles in the given interval:

Answer

\theta=-60^\circ,\ 60^\circ

Both satisfy the original equation: its two sides equal 1.

Solve for the ratio, then the angles

A quadratic can contain $\cos x$ instead of $x$ . Treat the entire cosine value as one unknown, then return to the angle.

Worked example

Solve $2\cos^2x-\cos x-1=0$ for $0^\circ\le x\le360^\circ$ .

Let $u=\cos x$ . Enter coefficients $2,-1,-1$ into the calculator's quadratic / degree-2 polynomial solver. The roots are $u=1$ and $u=-1/2$ , so write:

2u^2-u-1=0

(u-1)(2u+1)=0

The factor 2 is needed to keep the leading coefficient. Expanding checks the working. Now return to cosine:

\cos x=1\quad\text{or}\quad\cos x=-\frac12

The first gives 0° and 360°. The second gives 120° and 240°.

Answer

x=0^\circ,\ 120^\circ,\ 240^\circ,\ 360^\circ

Before pressing an inverse key, reject sine or cosine values outside $[-1,1]$ . Tangent has no such range restriction. Every valid ratio root needs its own angles.

Factor before dividing

For $2\sin^2x-\sin x=0$ , dividing by sine would lose every solution where sine is zero. Factor instead:

\sin x(2\sin x-1)=0

A product is zero when at least one factor is zero. Solve both $\sin x=0$ and $\sin x=1/2$ . On $0^\circ\le x\le360^\circ$ , that gives 0°, 30°, 150°, 180°, 360°.

Be careful with a squared substitution too. If $u=\sin^2x$ , then $0\le u\le1$ . A root $u=1/4$ gives $\sin x=\pm1/2$ , because both signs square to $1/4$ .

Your turn

Your turn9709/12 Oct/Nov 2021 Q1 [4 marks]

Solve for $-90^\circ<\theta<90^\circ$ :

2\cos\theta=7-\frac3{\cos\theta}

Show worked answer

Cosine cannot be zero in the original fraction. Multiply by cosine and bring all terms to one side:

2\cos^2\theta-7\cos\theta+3=0

The polynomial solver gives $1/2$ and 3. The supporting factorisation is:

(2\cos\theta-1)(\cos\theta-3)=0

Reject 3 because cosine cannot exceed 1. From $\cos\theta=1/2$ , use the two angles in the given interval:

Answer

\theta=-60^\circ,\ 60^\circ

Both satisfy the original equation: its two sides equal 1.