Quadratic Trigonometric Equations

Introduction

Quadratic trigonometric equations are equations where a trigonometric expression appears with degree $2$, just like a quadratic in algebra. For example, an equation such as $2\sin^2 x-3\sin x+1=0$ is quadratic in $\sin x$. students, this topic matters because it combines two big ideas in IB Mathematics Analysis and Approaches SL: algebraic solving and trigonometric reasoning. It is also closely connected to geometry, since trigonometry is used to study angles, triangles, circles, and periodic motion 🌟

Learning objectives

By the end of this lesson, you should be able to:

explain the meaning of a quadratic trigonometric equation,
solve equations such as $a\sin^2 x+b\sin x+c=0$ and $a\cos^2 x+b\cos x+c=0$,
interpret solutions using the allowed values of trigonometric functions,
connect these equations to graphs, identities, and real-world situations.

A key idea is that trigonometric values are restricted. For example, $\sin x$ and $\cos x$ must always lie between $-1$ and $1$. That restriction helps us decide which algebraic solutions are actually valid in a trigonometric context.

What makes an equation quadratic trigonometric?

A quadratic trigonometric equation is an equation that can be rewritten as a quadratic in a trigonometric function. Common forms include:

$a\sin^2 x+b\sin x+c=0$
$a\cos^2 x+b\cos x+c=0$
$a\tan^2 x+b\tan x+c=0$

Here, $a$, $b$, and $c$ are constants, and the variable is the angle $x$. The equation is called quadratic because the highest power of the trigonometric expression is $2$.

For example, in $\sin^2 x-3\sin x+2=0$, treat $\sin x$ like a single variable. Let $u=\sin x$. Then the equation becomes $u^2-3u+2=0$, which can be solved by factoring:

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

So $u=1$ or $u=2$. But $\sin x=2$ is impossible because sine values cannot be greater than $1$. Therefore the only valid equation is $\sin x=1$.

This shows a very important step: solving the algebra is not enough. You must also check whether the solution makes sense for the trig function.

Solving by substitution

The most common method is substitution. students, the process is similar to solving other quadratic equations:

Choose a trig function such as $\sin x$, $\cos x$, or $\tan x$.
Substitute a new variable, such as $u=\sin x$.
Solve the quadratic equation in $u$.
Convert the valid $u$ values back into trig equations.
Find the angle solutions in the required interval.

Example 1: solving in $[0^\circ,360^\circ)$

Solve $2\cos^2 x-3\cos x+1=0$.

Let $u=\cos x$. Then:

$2u^2-3u+1=0$

Factor:

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

So $u=\frac{1}{2}$ or $u=1$.

Now solve:

$\cos x=\frac{1}{2}$ gives $x=60^\circ$ and $x=300^\circ$
$\cos x=1$ gives $x=0^\circ$

So the solutions are:

$x=0^\circ,60^\circ,300^\circ$

Notice how the interval matters. If the domain were different, the list of answers could change.

Example 2: using tangent

Solve $\tan^2 x-4\tan x+3=0$ for $0^\circ\le x<180^\circ$.

Let $u=\tan x$. Then:

$u^2-4u+3=0$

Factor:

$(u-1)(u-3)=0$

So $\tan x=1$ or $\tan x=3$.

For $0^\circ\le x<180^\circ$:

$\tan x=1$ gives $x=45^\circ$
$\tan x=3$ gives one solution in Quadrant I and one in Quadrant III, but only the Quadrant I solution is in the interval, so $x\approx71.6^\circ$

Thus the solutions are:

$x=45^\circ,\;71.6^\circ\text{ (approximately)}$

Tangent is different from sine and cosine because it can take any real value, so there is no restriction like $-1\le \tan x\le 1$.

Using identities to rewrite the equation

Sometimes a quadratic trigonometric equation is written in a form that uses more than one trig function. In that case, identities help simplify the equation.

A very important identity is:

$\sin^2 x+\cos^2 x=1$

This can be used to rewrite an equation so that only one trig function appears.

Example 3: convert to one function

Solve $\sin^2 x+\sin x-2=0$.

Let $u=\sin x$. Then:

$u^2+u-2=0$

Factor:

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

So $u=-2$ or $u=1$.

Now check validity:

$\sin x=-2$ is impossible
$\sin x=1$ gives $x=90^\circ$ in $[0^\circ,360^\circ)$

So the answer is:

$x=90^\circ$

Example 4: using $\sin^2 x=1-\cos^2 x$

Solve $2\sin^2 x-\cos^2 x=1$.

Use $\sin^2 x=1-\cos^2 x$:

$2(1-\cos^2 x)-\cos^2 x=1$

Simplify:

$2-3\cos^2 x=1$

$3\cos^2 x=1$

$\cos^2 x=\frac{1}{3}$

So:

$\cos x=\pm\frac{1}{\sqrt{3}}$

Then find all angles in the required interval using a calculator or exact reasoning if possible.

This type of problem shows how algebraic identities and trigonometric graphs work together.

Graphs and number of solutions

Graphing helps you understand why a quadratic trig equation may have one, two, three, or more solutions in an interval. For example, if you solve:

$\sin^2 x-\sin x=0$

then factor:

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

So $\sin x=0$ or $\sin x=1$.

On the unit circle, $\sin x=0$ happens at $x=0^\circ,180^\circ,360^\circ$ in $[0^\circ,360^\circ]$, while $\sin x=1$ happens at $x=90^\circ$. The graph of $y=\sin^2 x-\sin x$ crosses the $x$-axis at those angles. This is a useful visual check ✅

Graphs also explain why some equations have no solutions. If a quadratic in $\sin x$ gives a root like $u=1.5$, that root cannot correspond to any angle because sine never reaches $1.5$.

Common mistakes to avoid

students, these are the most common errors students make:

forgetting to check whether algebraic roots are possible for a trig function,
missing solutions because only one angle was found instead of all angles in the interval,
mixing degrees and radians,
using the wrong identity,
stopping after factoring without converting back to $x$.

For example, if $\cos x=\frac{1}{2}$ in $[0^\circ,360^\circ)$, both $60^\circ$ and $300^\circ$ must be included. Leaving out one of them gives an incomplete solution.

Another careful point is interval notation. If the interval is $0\le x<2\pi$, then $0$ is included but $2\pi$ is not. In degrees, $0^\circ$ may be included while $360^\circ$ may not, depending on the interval.

Why this topic matters in Geometry and Trigonometry

Quadratic trigonometric equations connect directly to the wider course because they combine algebra, graphs, and geometric meaning. In geometry, trigonometric ratios describe side lengths and angles in right triangles, while the unit circle connects angles to coordinates. A quadratic trig equation often describes a geometric condition, such as a particular height, projection, or intersection point.

For example, if a model uses $h(t)=a\sin^2 t+b\sin t+c$, the solutions may represent times when a moving object reaches a certain height. In a geometry setting, equations involving $\sin x$ or $\cos x$ may come from triangle relationships or circle problems. So this topic is not isolated; it supports problem solving across the whole Geometry and Trigonometry unit.

Conclusion

Quadratic trigonometric equations are solved by combining algebraic methods with trigonometric knowledge. The main strategy is to rewrite the equation as a quadratic in one trig function, solve the quadratic, and then keep only the values that are possible for that function. students, the key skill is checking both the algebra and the geometry. This topic strengthens your understanding of identities, intervals, graphs, and real-world modelling, making it an important part of IB Mathematics Analysis and Approaches SL 📘

Study Notes

A quadratic trigonometric equation has a trig expression with power $2$, such as $a\sin^2 x+b\sin x+c=0$.
Use substitution, such as $u=\sin x$, to turn the equation into an ordinary quadratic.
Solve the quadratic, then check that the roots are valid for the trig function.
Remember $-1\le \sin x\le 1$ and $-1\le \cos x\le 1$.
Tangent can take any real value, so $\tan x$ has no such range restriction.
Identities like $\sin^2 x+\cos^2 x=1$ help rewrite equations into one trig function.
Always find all solutions in the given interval, such as $[0^\circ,360^\circ)$ or $[0,2\pi)$.
Graphs and the unit circle are useful for checking the number and location of solutions.
This topic connects algebraic solving with geometry, angles, and modelling.