Trig equations: elementary, linear, homogeneous

In Year 4 at the liceo scientifico (the Italian science-focused high school), trigonometric equations come right after the addition and double-angle formulas, usually between September and October. The test that includes them falls in November, so you have five to eight weeks to turn them from a “list of cases to remember” into a method that always starts from the same point: recognising what type of equation it is. Equations divide into five types, and for each one there is a fixed move.

This post is the guide to the solving technique. You will find the basic formulas in trigonometry for the maturità, the picture of the whole test in trigonometry in Year 4, and the full programme of the year in Year 4: the formula sheet. Here we care about only one thing: given a question, how to reach the solutions without losing any and without inventing any.

The overview: which type of equation is it

Before calculating, look at the terms and classify.

Elementary equations

These are $\sin x=a$, $\cos x=a$, $\tan x=a$. They have a solution if $-1\le a\le 1$ for sine and cosine (the tangent allows any $a$). With $\alpha$ the principal value:

Sine and cosine have two families because on the unit circle there are two points with the same ordinate (or the same abscissa). The tangent has only one and its period is $\pi$.

Three special cases not to treat with the general formula, because they can be read directly from the circle: $\sin x=1$ gives $x=\frac{\pi }{2}+2k\pi$, $\sin x=-1$ gives $x=-\frac{\pi }{2}+2k\pi$, $\sin x=0$ gives $x=k\pi$. Here there is only one family, not two: the two coincide.

If the argument is not $x$ the procedure is the same, but at the end you solve for $x$ dividing the period too. From $\sin 2x=\frac{1}{2}$ you get $2x=\frac{\pi }{6}+2k\pi$ or $2x=\frac{5\pi }{6}+2k\pi$, so

The period becomes $\pi$: it is the mistake I see most often, and you will find it again among the typical mistakes further down.

Equations reducible to elementary ones

You sometimes find $\sin 3x=\cos x$ or $\sin x=\sin \left(2x-\frac{\pi }{4}\right)$. With the same sine on both sides you apply the two families directly, equating the arguments. When the functions are different, you use an associated angle to bring them to the same one:

So $\sin 3x=\cos x$ becomes $\sin 3x=\sin \left(\frac{\pi }{2}-x\right)$, and from here the two families: $3x=\frac{\pi }{2}-x+2k\pi$ that is $x=\frac{\pi }{8}+\frac{k\pi }{2}$, or $3x=\pi -\frac{\pi }{2}+x+2k\pi$ that is $x=\frac{\pi }{4}+k\pi$.

After dividing by the coefficient of $x$, the period changes: in the first family $\frac{2k\pi }{4}=\frac{k\pi }{2}$. If you get $2k\pi$ after dividing by 4, you have lost solutions.

Quadratic in sine or cosine

You see ${\sin }^{2}x$ and $\sin x$ (or the corresponding ones in cosine). If both functions appear, you use the identity ${\sin }^{2}x+{\cos }^{2}x=1$ to rewrite everything in one. Then you set $t=\sin x$ and solve the trinomial. Watch the constraint: $t$ must lie in $[-1,1]$, and roots outside this interval are discarded with a line of comment. For each remaining root you fall back on an elementary equation.

Linear equations: $a\sin x+b\cos x=c$

This is the type that makes the difference in November, because there are two methods and it pays to know both.

First the existence condition for solutions. The left-hand side, up to a constant, is a sine: it has maximum $\sqrt{a^2+b^2}$ and minimum $-\sqrt{a^2+b^2}$. The equation is solvable only if $|c|\le \sqrt{a^2+b^2}$. So $\sin x+\cos x=2$ is impossible, because $\sqrt{2}<2$: you know it before doing any calculation.

The auxiliary angle method. You divide by $\sqrt{a^2+b^2}$ and look for an angle $\phi$ such that $\cos \phi =\frac{a}{\sqrt{a^2+b^2}}$ and $\sin \phi =\frac{b}{\sqrt{a^2+b^2}}$. Then the left-hand side is $\sin (x+\phi )$ and the equation becomes elementary. With coefficients such as $1$ and $\sqrt{3}$, or $1$ and $1$, the angle $\phi$ is a standard one. When it is not, you leave it expressed with the arctangent.

The parametric method. You set $t=\tan \frac{x}{2}$, so that

and you get a quadratic equation in $t$. It has a hidden cost: the substitution does not exist for $x=\pi +2k\pi$, where $\tan \frac{x}{2}$ is not defined. That angle has to be checked separately, by substituting it into the starting equation. Exercise 2 shows what happens if you forget to do it.

Homogeneous equations of the second degree

The form is

Every term has degree 2 taken together in sine and cosine. The move: divide everything by ${\cos }^{2}x$, obtaining a quadratic equation in $\tan x$:

But dividing by ${\cos }^{2}x$ is allowed only if $\cos x\ne 0$. The check is one line: substitute $\cos x=0$ (that is, ${\sin }^{2}x=1$) into the equation. If $a=0$ holds, then $x=\frac{\pi }{2}+k\pi$ is a solution and has to be added to those found with the tangent. With $a\ne 0$ it is not.

If the equation has a constant term, for example ${\sin }^{2}x-\sin x\cos x=1$, the trick is to write $1={\sin }^{2}x+{\cos }^{2}x$: the constant term becomes a homogeneous term and you fall back on the previous case. If instead the term in ${\cos }^{2}x$ is missing (that is, $c=0$), there is no need to divide: you factor out $\sin x$ or $\cos x$ and get a product.

Three worked examples

1. Quadratic, with an extra family

With ${\cos }^{2}x=1-{\sin }^{2}x$ the equation becomes $2-2{\sin }^{2}x+3\sin x-3=0$, that is

Setting $t=\sin x$, you have $2t^2-3t+1=0$ and therefore $t=1$ or $t=\frac{1}{2}$, both in the allowed interval. The solutions:

Check. For $x=\frac{\pi }{2}$: $0+3-3=0$. For $x=\frac{\pi }{6}$, with ${\cos }^{2}x=\frac{3}{4}$ and $\sin x=\frac{1}{2}$: $\frac{3}{2}+\frac{3}{2}-3=0$. For $x=\frac{5\pi }{6}$ the values of the sine and of the squared cosine are the same, so the same calculation holds. The case $\sin x=1$ has a single family, not two.

2. Linear with both methods, and the lost point

Existence check: $\sqrt{1+1}=\sqrt{2}\ge 1$, so there are solutions.

Auxiliary angle. Dividing by $\sqrt{2}$ you get $\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x=\frac{1}{\sqrt{2}}$, and the left-hand side is $\sin \left(x-\frac{\pi }{4}\right)$:

So $x-\frac{\pi }{4}=\frac{\pi }{4}+2k\pi$ or $x-\frac{\pi }{4}=\frac{3\pi }{4}+2k\pi$, that is

Check: for $x=\frac{\pi }{2}$ you have $1-0=1$; for $x=\pi$ you have $0-(-1)=1$.

Parametric. With $t=\tan \frac{x}{2}$ the equation becomes $\frac{2t}{1+t^2}-\frac{1-t^2}{1+t^2}=1$. Multiplying by $1+t^2$:

and $\tan \frac{x}{2}=1$ gives $x=\frac{\pi }{2}+2k\pi$. $x=\pi +2k\pi$ is missing, which is exactly the angle excluded by the substitution: if you stop here you lose half the solutions. It has to be substituted separately, and as we saw above it is a solution. That is why, when the coefficients allow it, I prefer the auxiliary angle: it hides no angle.

3. Homogeneous, with the check on the division

Check: if $\cos x=0$ then ${\sin }^{2}x=1$ and the left-hand side equals $2\ne 0$, so $x=\frac{\pi }{2}+k\pi$ is not a solution and you can divide by ${\cos }^{2}x$:

With $t=\tan x$ you have $2t^2-t-1=0$, that is $(2t+1)(t-1)=0$, and $t=1$ or $t=-\frac{1}{2}$. The solutions:

Check for the second: with $\tan x=-\frac{1}{2}$ you have ${\sin }^{2}x=\frac{t^2}{1+t^2}=\frac{1}{5}$, ${\cos }^{2}x=\frac{4}{5}$ and $\sin x\cos x=\frac{t}{1+t^2}=-\frac{2}{5}$. The left-hand side equals $\frac{2}{5}+\frac{2}{5}-\frac{4}{5}=0$. For the first, with $x=\frac{\pi }{4}$: $2\cdot \frac{1}{2}-\frac{1}{2}-\frac{1}{2}=0$.

If you want to practise with a wider range of cases of this kind, there is the trigonometry practice test, which follows the structure of Year 4 tests. For the set-up of all the calculation side you will find the rest of the work in maths.

The mistakes I see most often

Forgetting the period. Writing a single family for a sine or a cosine, or using $2k\pi$ for a tangent. After dividing the argument by a coefficient, the period is divided in the same way. From $3x=\cdots +2k\pi$ you get $x=\cdots +\frac{2k\pi }{3}$.

Dividing by $\cos x$ (or by ${\cos }^{2}x$) without checking. With the equation $\sin x\cos x=\cos x$, dividing by $\cos x$ leads to $\sin x=1$ and wipes out all the solutions of $\cos x=0$. The right way is to factor: $\cos x(\sin x-1)=0$, which gives $\cos x=0$ or $\sin x=1$. Never cancel a factor that can be zero: you always bring everything to the left and factor.

Extraneous solutions. They appear in two situations: if you square (the solutions of the squared equation have to be substituted back into the original), if the equation contains a $\tan x$ or a denominator (the solutions with $\cos x=0$ are discarded). A final substitution, even on just one value per family, catches almost all of them.

Roots of $t$ outside $[-1,1]$. In trinomials in $\sin x$ or $\cos x$ there is nothing to look for when $|t|>1$: you write “impossible” and move on, otherwise you attempt arcsines of undefined values.

Parametric without checking $\pi$. It happens to almost everyone the first time. You saw it in exercise 2.

How I get you ready for November

The November test almost always contains a homogeneous equation, a linear one and a quadratic one, often with an argument other than $x$ to see whether you can handle the period. Good training is to solve one equation per type, always check one solution per family and write a line of verification like the ones in the exercises above: the teacher reads them, and it saves you from slips.

If you reach this chapter with doubts about the addition formulas or the basic identities, it is better to sort those out first: equations are only their application. In lessons we work exactly that way, starting from last year’s tests and the mistakes that come up most often. If you want a free introductory call to see where you need to work before the test, you can book it from contact.