Solve linear and quadratic trigonometric equations

Ontario Grade 12 Mathematics

Trigonometric Functions

Find every angle that satisfies an equation on a stated interval.

A trigonometric equation contains a ratio such as sine or cosine. Its solutions are angles. A single ratio value can match more than one angle, so solving the algebra is only part of the work. You must find every matching angle in the interval given in the question. This lesson reviews the angle facts needed and applies them to linear and quadratic equations.

What you will learn

Recognize linear and quadratic trigonometric equations.
Solve a linear trigonometric equation by isolating the ratio and finding all matching angles.
Solve a quadratic trigonometric equation by finding possible ratio values and then the angles.
Check solutions in the original equation and the stated interval.

Prerequisite bridge: ratios, angles, and intervals

An equation says that two expressions are equal. Solving it means finding values that make the statement true. In a trigonometric equation, the unknown is usually an angle, written as

x

or

\theta

. The trigonometric ratio is the sine or cosine of that angle.

Sine and cosine can be pictured using the unit circle, a circle with radius

1

. At angle

x

, the point on the circle has coordinates

(\cos x,\sin x)

. Cosine is the horizontal coordinate, and sine is the vertical coordinate. Since a point on this circle cannot be more than

1

unit from its centre in either direction, sine and cosine values are between

-1

and

1

.

Radians are a way to measure angles. One complete turn is

2\pi

radians. An interval states which angles to include. For example,

[0,2\pi)

includes

0

but excludes

2\pi

. The two endpoints point in the same direction, but the interval includes only one of them.

A reference angle is the acute angle between an angle's terminal arm and the horizontal axis. It helps connect angles in different parts of the circle to familiar positive angle values. The sign of the sine or cosine tells you which parts of the circle can contain solutions.

-1\leq\sin x\leq1,\qquad -1\leq\cos x\leq1

The unknown is an angle, while the value of its trigonometric ratio is a number.
The unit circle connects sine and cosine values to angles.
Follow the interval, including its endpoint rules.

Linear equations: isolate the ratio, then find angles

A linear trigonometric equation has a trigonometric ratio to the first power. For example,

3\sin x-1=0

is linear in

\sin x

. Solve the ordinary algebra first: undo addition or subtraction, then undo multiplication or division, until the ratio is alone. Each algebra step must preserve equality.

For example,

\sin x=\frac{1}{2}

on

[0,2\pi)

means the vertical coordinate on the unit circle is

\frac{1}{2}

. It occurs at

\frac{\pi}{6}

and

\frac{5\pi}{6}

. These are different angles with the same sine value. Giving just one angle would leave out a solution.

The unit circle or familiar angle values can help you find the angles. For a value that is not familiar, an inverse trigonometric function can give a starting angle. That starting value may not list every solution. Use the ratio's sign and the stated interval to locate all matching angles. An inverse value is a tool for finding angles, not a replacement for checking the full interval.

\sin x=\frac{1}{2}\;\Longrightarrow\;x=\frac{\pi}{6},\frac{5\pi}{6}\text{ on }[0,2\pi)

Rearrange first so the ratio is isolated.
One ratio value may match multiple angles in the interval.
An inverse trigonometric value may give only one of the required angles.

Quadratic equations: solve for ratio values

A quadratic trigonometric equation contains a squared ratio or another quadratic expression in a ratio. For example,

a\sin^2 x+b\sin x+c=0

is quadratic in

\sin x

. Treat

\sin x

as the quantity being squared, much as you would treat a letter in an ordinary quadratic equation. Factor when possible, or use a quadratic method to find the possible sine values.

Each factor gives a possible value for the ratio. These are not yet angle solutions. For every possible sine or cosine value, find all angles in the interval that have that value. Before doing so, check the range: a value outside the interval from

-1

to

1

cannot be the sine or cosine of a real angle.

Keep the algebra and angle-finding steps separate. The algebra solves for possible ratio values. The unit circle or angle values identify the angles. Finally, substitute the ratio values into the original equation and check that the resulting angles belong to the stated interval.

a\sin^2 x+b\sin x+c=0

Solve the quadratic for ratio values before finding angles.
Discard values outside the range from $-1$ to $1$ .
Find every matching angle and check it in the original equation.

A routine for solving and checking

Start by identifying the trigonometric ratio and the interval. For a linear equation, isolate the ratio. For a quadratic equation, solve the quadratic for possible ratio values. Do not stop when you have values for sine or cosine: the question asks for angles.

Use familiar angle values, the unit circle, or an inverse trigonometric value to locate the angles. Consider the sign of the ratio and check each possible angle against the interval. Pay close attention to square brackets and round brackets because they show whether an endpoint is included.

Check each candidate in two ways. Confirm that its angle belongs to the interval, and confirm that its ratio value makes the original equation true. Write the final solutions as angles. If there are no angles in the interval that satisfy the equation, state that there are no solutions in that interval.

Separate solving for ratio values from finding angles.
Check each answer in the equation and the interval.
Report the angles as the final solutions.

Familiar unit-circle values

Angle	$\sin x$	$\cos x$
$\frac{\pi}{6}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$
$\frac{\pi}{4}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$
$\frac{\pi}{3}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$
$\frac{5\pi}{6}$	$\frac{1}{2}$	$-\frac{\sqrt{3}}{2}$
$\frac{7\pi}{6}$	$-\frac{1}{2}$	$-\frac{\sqrt{3}}{2}$
$\frac{11\pi}{6}$	$-\frac{1}{2}$	$\frac{\sqrt{3}}{2}$

Worked example

Solve a quadratic trigonometric equation

Solve

2\sin^2 x-\sin x-1=0

for x∈[0,2\pi).

Factor the quadratic
Treat $\sin x$ as the quantity being squared. The factors multiply to the original quadratic, so their product is zero when at least one factor is zero.
$(2\sin x+1)(\sin x-1)=0$
Find the possible sine values
Set each factor equal to zero and solve. The two possible values of sine are $-\frac{1}{2}$ and $1$ . These are ratio values, not angle solutions. In other words, $\sin x=-\frac{1}{2}$ or $\sin x=1$ .
Find the angles
On $[0,2\pi)$ , sine equals $-\frac{1}{2}$ at two angles below the horizontal axis. Sine equals $1$ at the top of the unit circle. All three angles are in the interval.
$x=\frac{7\pi}{6},\frac{11\pi}{6},\frac{\pi}{2}$
Check the candidates
Substituting either possible sine value into the original quadratic gives zero. Each listed angle also belongs to the stated interval.
$2\left(-\frac{1}{2}\right)^2-\left(-\frac{1}{2}\right)-1=0,\qquad 2(1)^2-1-1=0$

Answer:

x=\frac{\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6}

Check: The three angles have sine values

1

,

-\frac{1}{2}

, and

-\frac{1}{2}

. Each value makes the original equation true, and every angle is in

[0,2\pi)

.

Common mistakes and how to avoid them

Reporting only the angle from an inverse trigonometric function.

Correction: Use the unit circle and the interval to find every angle with the required ratio value.

Treating a solution of the quadratic in

\sin x

as an angle.

Correction: That solution is a possible sine value. Find the angles whose sine equals that value.

Keeping a sine or cosine value outside the range from

-1

to

1

.

Correction: Reject that value because no real angle can have it as its sine or cosine.

Including an excluded endpoint or overlooking the interval.

Correction: Check each angle against the interval notation. For

[0,2\pi)

, include

0

and exclude

2\pi

.

Lesson summary

For a linear equation, isolate the trigonometric ratio and find all matching angles.
For a quadratic equation, solve for possible ratio values first, then find the corresponding angles.
Check every solution in the original equation and against the interval.

Check your understanding

Question 1

Solve

\cos x=0

for x∈[0,2\pi).

$x=\frac{\pi}{2},\frac{3\pi}{2}$
$x=0,\pi$
$x=\frac{\pi}{2}$ only
$x=\frac{\pi}{2},\frac{3\pi}{2},2\pi$

Show answer and explanation

x=\frac{\pi}{2},\frac{3\pi}{2}

Cosine is the horizontal coordinate on the unit circle. It is zero at the top and bottom points, giving

\frac{\pi}{2}

and

\frac{3\pi}{2}

. The endpoint

2\pi

is excluded and is not a solution.

Question 2

Solve

\sin^2 x-1=0

for x∈[0,2\pi).

$x=0,\pi$
$x=\frac{\pi}{2},\frac{3\pi}{2}$
$x=\frac{\pi}{2}$ only
$x=\frac{\pi}{2},\frac{3\pi}{2},2\pi$

Show answer and explanation

x=\frac{\pi}{2},\frac{3\pi}{2}

Factoring gives

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

, so

\sin x=1

or

\sin x=-1

. These occur at

\frac{\pi}{2}

and

\frac{3\pi}{2}

in the stated interval.

Key terms

Trigonometric equation: An equation containing a ratio such as sine or cosine.
Linear trigonometric equation: An equation in which the trigonometric ratio appears to the first power.
Quadratic trigonometric equation: An equation in which a trigonometric ratio appears squared or in a quadratic expression.
Reference angle: The acute angle between an angle's terminal arm and the horizontal axis.
Interval: A range that states which angles may be solutions.

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Published by DoAssignment. This reviewed lesson follows Ontario Grade 12 Mathematics (MHF4U), expectation B3.4. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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