P2 6.2 - Solving trigonometric equations | Edexcel A-Level International Mathematics

Find every solution of a trigonometric equation in a stated interval, including shifted angles, multiple angles and quadratic expressions. You will use the interval throughout the calculation and check that algebra has neither lost valid solutions nor introduced undefined ones.

All solutions in a specified interval

Why can $\sin x=1/2$ have two solutions in one interval and four in another? A calculator gives one inverse-trig value, but the graph repeats: the interval determines how many answers to keep.

In one full turn, sine is positive in quadrants I and II. The reference angle $30^\circ$ therefore gives $x=30^\circ$ and $150^\circ$ . Adding or subtracting $360^\circ$ gives further solutions with the same sine.

[DIAGRAM: asset_name: Sine solutions in a wider interval; asset_slug: sine-interval-solutions; caption: Four intersections lie in the interval from minus 360 degrees inclusive to 360 degrees exclusive.; description: The sine graph from minus 360 to 360 degrees intersects y equal to one half at minus 330, minus 210, 30 and 150 degrees. The four intersection points are labelled, showing two periods rather than only one calculator answer.]
Four intersections lie in the interval from minus 360 degrees inclusive to 360 degrees exclusive.

Use the graph or quadrant signs to find the other angles. For a positive acute reference angle $\alpha$ , the usual locations in one turn are:

Equation	Solutions in one turn, when the stated value is positive
$\sin x=\sin\alpha$	$\alpha,\ 180^\circ-\alpha$
$\cos x=\cos\alpha$	$\alpha,\ 360^\circ-\alpha$
$\tan x=\tan\alpha$	$\alpha,\ 180^\circ+\alpha$

For a negative value, choose the quadrants with the required sign. Sine and cosine repeat every $360^\circ$ , or $2\pi$ radians; tangent repeats every $180^\circ$ , or $\pi$ radians. Add or subtract complete periods until every angle in the specified interval is included.

Worked example 1: one full turn

Question. Solve $\sin x=1/2$ for $0^\circ\le x<360^\circ$ .

Plan. Find the reference angle, then use the two quadrants where sine is positive.

\alpha=\sin^{-1}(1/2)=30^\circ, \qquad x=30^\circ,\ 180^\circ-30^\circ.

Answer. $x=30^\circ,\ 150^\circ$ .

Check. Both angles have sine $1/2$ . The sine graph crosses this positive level twice in the stated full turn.

Change only the interval: include a second full turn on the negative side of zero.

Worked example 2: two full turns

Question. Solve $\sin x=1/2$ for $-360^\circ\le x<360^\circ$ .

Plan. Start with the two solutions from one turn, then subtract $360^\circ$ and retain all values in range.

\begin{aligned} 30^\circ-360^\circ&=-330^\circ\\ 150^\circ-360^\circ&=-210^\circ \end{aligned}

Answer. $x=-330^\circ,\ -210^\circ,\ 30^\circ,\ 150^\circ$ .

Check. Adding $360^\circ$ to either negative answer produces its positive partner, with the same sine. The graph above shows exactly four intersections.

Endpoints matter. In $0^\circ\le x<360^\circ$ , $0^\circ$ is allowed and $360^\circ$ is excluded. For example, $\sin x=0$ has answers $0^\circ,180^\circ$ in that interval, but also $360^\circ$ if the upper inequality is changed to $\le$ . Special axis values need not give two different answers per turn.

Set your calculator to the angle unit requested. Keep exact standard angles as fractions of $\pi$ when radians are required.

Shifted and multiple angles

When the function contains an angle such as $2x+30^\circ$ , solve for that whole angle first. Its interval changes too.

If $-180^\circ<x<180^\circ$ , then $-330^\circ<2x+30^\circ<390^\circ$ . This wider interval contains more cycles than the interval for $x$ . Solving only for an inverse-trig principal value will miss answers.

Problem-solving

Set $u$ equal to the whole angle. Transform both interval endpoints, find every allowed $u$ , then undo the shift and multiplication in reverse order.

This method treats the expression inside the function as one angle. It does not require expanding a trigonometric function of a sum.

Worked example 3: a shifted angle in degrees

Question. Solve $\cos(x+30^\circ)=1/2$ for $-180^\circ<x<180^\circ$ .

Plan. Set $u=x+30^\circ$ , solve for $u$ in its shifted interval, then subtract $30^\circ$ .

-150^\circ<u<210^\circ.

Cosine is $1/2$ at $u=-60^\circ,60^\circ$ in this interval. Hence

x=-60^\circ-30^\circ,\quad 60^\circ-30^\circ.

Answer. $x=-90^\circ,\ 30^\circ$ .

Check. Their original angles are $-60^\circ$ and $60^\circ$ , both with cosine $1/2$ ; both $x$ -values lie strictly within the original interval.

Change only $x$ to $2x$ inside the angle. The transformed interval is now twice as long.

Worked example 4: a multiple angle in degrees

Question. Solve $\cos(2x+30^\circ)=1/2$ for $-180^\circ<x<180^\circ$ .

Plan. Set $u=2x+30^\circ$ , list every cosine solution in the transformed interval, then use $x=(u-30^\circ)/2$ .

-330^\circ<u<390^\circ, \qquad u=-300^\circ,-60^\circ,60^\circ,300^\circ.

Therefore

\begin{aligned} u=-300^\circ&\ \Longrightarrow\ x=-165^\circ\\ u=-60^\circ&\ \Longrightarrow\ x=-45^\circ\\ u=60^\circ&\ \Longrightarrow\ x=15^\circ\\ u=300^\circ&\ \Longrightarrow\ x=135^\circ \end{aligned}

Answer. $x=-165^\circ,\ -45^\circ,\ 15^\circ,\ 135^\circ$ .

Check. Substitution gives whole angles $-300^\circ,-60^\circ,60^\circ,300^\circ$ , all with cosine $1/2$ . All four $x$ -values satisfy the strict interval.

Radians use the same process. Tangent has period $\pi$ , which lets you list its whole-angle solutions especially efficiently.

Worked example 5: a tangent multiple angle

Question. Solve $\tan2x=1$ for $0\le x<\pi$ , giving exact answers.

Plan. Use $u=2x$ , so $0\le u<2\pi$ , and add periods of $\pi$ to $\pi/4$ .

u=\frac\pi4,\ \frac{5\pi}4 \quad\Longrightarrow\quad x=\frac\pi8,\ \frac{5\pi}8.

Answer. $x=\pi/8,\ 5\pi/8$ .

Check. Doubling gives $\pi/4$ and $5\pi/4$ , whose tangents are both $1$ . Two tangent periods occur in the interval for $u$ .

Change only the shift in the whole angle, from zero to $\pi/6$ .

Worked example 6: a shifted tangent multiple angle

Question. Solve $\tan(2x+\pi/6)=1$ for $0\le x<\pi$ , giving exact answers.

Plan. Transform the interval before choosing the tangent solutions.

\frac\pi6\le u=2x+\frac\pi6<\frac{13\pi}6.

The values $u=\pi/4,5\pi/4$ are in this interval; the next, $9\pi/4$ , is above it. Therefore

x=\frac{\pi/4-\pi/6}{2}=\frac\pi{24}, \qquad x=\frac{5\pi/4-\pi/6}{2}=\frac{13\pi}{24}.

Answer. $x=\pi/24,\ 13\pi/24$ .

Check. Substitution gives $2x+\pi/6=\pi/4,5\pi/4$ , respectively, so both satisfy the original equation and interval.

The shift is undone before division by the multiplier. Writing $x=u/2-\pi/6$ here would subtract too much; it must be $x=(u-\pi/6)/2$ .

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P2 6.2 - Solving trigonometric equations

All solutions in a specified interval

Shifted and multiple angles

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