E7 - Solving trigonometric equations | AQA A-Level Mathematics

Find every solution in a stated interval, using symmetry, periodicity and algebra. Transform changed arguments, solve quadratics in trigonometric functions, and check that admissibility restrictions or cancelled factors have not altered the solution set.

Branches and interval endpoints

Why can one inverse-sine button miss most of an equation's answers? For $\sin x=-\sqrt3/2$ , a calculator gives $-\pi/3$ . Yet the interval $0\le x<2\pi$ contains two answers, and that principal value is neither of them. The calculator answers its inverse-function question; you still have to answer the interval question.

You already know

Sine and cosine repeat after $2\pi$ radians or $360^\circ$ . Tangent repeats after $\pi$ radians or $180^\circ$ . Use the angle unit stated in the question, and remember that an open endpoint is excluded.

Use symmetry to find the other branch, then add whole periods. The two branches coincide at sine or cosine values $1$ and $-1$ ; list each input only once.

All branches before interval filtering

For integer $k$ , in radians,

\begin{aligned} \sin u=c&:\quad u=a+2k\pi\text{ or }u=\pi-a+2k\pi\quad(a=\arcsin c)\\ \cos u=c&:\quad u=\pm b+2k\pi\quad(b=\arccos c)\\ \tan u=c&:\quad u=d+k\pi\quad(d=\arctan c) \end{aligned}

Sine and cosine require $-1\le c\le1$ . Tangent permits any real $c$ , while excluding its undefined inputs. In degrees replace $\pi$ by $180^\circ$ and $2\pi$ by $360^\circ$ .

Not in the formula booklet: learn it.

These branches organise the search; the requested interval decides which members survive.

Worked example 1: a negative sine

Question. Solve $\sin x=-\sqrt3/2$ for $0\le x<2\pi$ .

Plan. Use the reference angle and the two negative-sine quadrants.

\begin{aligned} \text{reference angle}&=\frac\pi3 &&\text{exact sine value}\\ x&=\pi+\frac\pi3,\ 2\pi-\frac\pi3 &&\text{quadrants III and IV}\\ x&=\frac{4\pi}3,\frac{5\pi}3 &&\text{simplify} \end{aligned}

Why this step? Both unit-circle points with the required negative height occur inside the interval.

Answer. $x=4\pi/3,5\pi/3$ .

Check. Sine at each is $-\sqrt3/2$ ; neither endpoint has that sine value.

Change only the right-hand side to $-1/2$ . The reference angle changes, but the required quadrants do not.

Worked example 2: a different reference angle

Question. Solve $\sin x=-1/2$ for $0\le x<2\pi$ .

Plan. Use reference angle $\pi/6$ in the same two quadrants.

\begin{aligned} x&=\pi+\frac\pi6,\ 2\pi-\frac\pi6 &&\text{negative sine branches}\\ x&=\frac{7\pi}6,\frac{11\pi}6 &&\text{filter to the interval} \end{aligned}

Why this step? A negative principal value can be moved into the interval by a full turn; it must not be discarded without generating its equivalents.

Answer. $x=7\pi/6,11\pi/6$ .

Check. The angles have equal sine, and both lie strictly between $\pi$ and $2\pi$ .

For a degree example, $\tan x=1$ on $-180^\circ\le x\le180^\circ$ gives $x=45^\circ+180^\circ k$ , hence $-135^\circ,45^\circ$ . The period is a half-turn, not a full turn.

Watch out

Wrong: $\sin x=-1/2\Rightarrow x=-\pi/6$ only. That is one principal inverse value. Right: generate both branches and periods, then filter to the stated interval.

Keep the interval and negative sine, and change only its magnitude again.

Multiples and shifted arguments

The whole argument needs its own interval. When $u=mx+c$ , with $m\ne0$ , transform the endpoints before searching for $u$ , then recover $x=(u-c)/m$ .

For $m>0$ , the endpoint order is preserved. For $m<0$ , reverse it and carry each endpoint's open or closed status with it: for example, $0\le x<\pi$ becomes $-2\pi<u\le0$ when $u=-2x$ . These linear transformations are not formula-booklet entries.

Worked example 3: a double argument

Question. Solve $\cos(2x-\pi/3)=1/2$ for $0\le x<2\pi$ .

Plan. Transform the interval and use both cosine branches over it.

\begin{aligned} u&=2x-\pi/3\quad\Rightarrow\quad-\pi/3\le u<11\pi/3 &&\text{transform endpoints}\\ u&=\pm\pi/3+2k\pi &&\text{cosine branches}\\ u&=-\pi/3,\pi/3,5\pi/3,7\pi/3 &&\text{permitted values}\\ x&=0,\pi/3,\pi,4\pi/3 &&\text{use }x=(u+\pi/3)/2 \end{aligned}

Why this step? The argument covers two turns; searching just one turn would lose two answers.

Answer. $x=0,\pi/3,\pi,4\pi/3$ .

Check. The next argument value $11\pi/3$ would give $x=2\pi$ , the excluded endpoint. All listed arguments have cosine $1/2$ .

Change only the argument multiplier from $2$ to $3$ . The argument now covers three turns.

Worked example 4: a triple argument

Question. Solve $\cos(3x-\pi/3)=1/2$ for $0\le x<2\pi$ .

Plan. Extend the argument interval and then divide by the new multiplier.

\begin{aligned} -\pi/3&\le u=3x-\pi/3<17\pi/3 &&\text{transform endpoints}\\ u&=-\pi/3,\pi/3,5\pi/3,7\pi/3,11\pi/3,13\pi/3 &&\text{both branches}\\ x&=0,2\pi/9,2\pi/3,8\pi/9,4\pi/3,14\pi/9 &&\text{restore }x \end{aligned}

Why this step? Each full argument turn contributes two cosine solutions here, so three turns contribute six.

Answer. The six values in the final line are the complete solution set.

Check. The next argument $17\pi/3$ again gives the excluded input $2\pi$ .

A useful surprise is that $\sin(nx)=0$ has exactly $2n$ distinct solutions on $0\le x<2\pi$ for every positive integer $n$ . Indeed, $nx=k\pi$ gives $x=k\pi/n$ , with integers $k=0,\ldots,2n-1$ . A simple change of frequency creates as many roots as you like. This counting result is derived, not separately in the formula booklet.

Watch out

Wrong: after solving $\cos u=1/2$ , restrict $u$ to $[0,2\pi)$ regardless of its definition. Right: transform the original interval for the whole argument first, including its endpoint status.

Change just the multiplier again, to $4$ .

Quadratics and permissible function values

Treat a quadratic in one trigonometric function as an algebra problem first. The substitution $s=\sin x$ or $c=\cos x$ converts it into an ordinary quadratic, but only roots in $[-1,1]$ can become real angles.

This is the same substitution-and-admissibility idea used with algebraic functions. Factorisation, completing the square or the quadratic formula can solve the temporary quadratic. Its roots are function values, not yet the requested inputs.

Worked example 5: two admissible sine values

Question. Solve $2\sin^2x+\sin x-1=0$ for $0\le x<2\pi$ .

Plan. Solve for $s=\sin x$ , check its range, then solve both sine equations.

\begin{aligned} 2s^2+s-1&=(2s-1)(s+1)=0 &&\text{factorise}\\ s&=\frac12\text{ or }-1 &&\text{both admissible}\\ \sin x=\frac12&\Rightarrow x=\frac\pi6,\frac{5\pi}6 &&\text{two branches}\\ \sin x=-1&\Rightarrow x=\frac{3\pi}2 &&\text{one distinct extreme input} \end{aligned}

Why this step? The two branches for sine $-1$ coincide, so they do not create two different answers.

Answer. $x=\pi/6,5\pi/6,3\pi/2$ .

Check. Both $s=1/2$ and $s=-1$ make the original quadratic zero.

Change only the second quadratic factor from $s+1$ to $s+3$ . One algebraic root now falls outside sine's range.

Worked example 6: rejecting an impossible value

Question. Solve $2\sin^2x+5\sin x-3=0$ for $0\le x<2\pi$ .

Plan. Factor in $s=\sin x$ and test the two resulting values before finding angles.

\begin{aligned} 2s^2+5s-3&=(2s-1)(s+3)=0 &&\text{factorise}\\ s&=\frac12\text{ or }-3 &&\text{quadratic roots}\\ s=-3&\text{ is rejected} &&\text{sine lies in }[-1,1]\\ x&=\frac\pi6,\frac{5\pi}6 &&\sin x=1/2 \end{aligned}

Why this step? An algebraic solution for the substitute may not correspond to any real input of the original function.

Answer. $x=\pi/6,5\pi/6$ .

Check. Substituting $s=1/2$ gives $1/2+5/2-3=0$ .

The same method works for cosine. For example, $2\cos^2x+5\cos x-3=0$ has cosine values $1/2,-3$ ; only $1/2$ survives, giving $\pi/3,5\pi/3$ on this half-open interval. This is an application of the learned identities and quadratic methods, not a new booklet formula.

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E7 - Solving trigonometric equations

Branches and interval endpoints

Multiples and shifted arguments

Quadratics and permissible function values

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