Solve first- and second-degree trigonometric equations with stated domains

Alberta Grade 12 Mathematics

Trigonometry

TR5: Find every solution in the stated domain

A trigonometric equation asks which angles make a statement involving a trigonometric ratio true. The domain tells you which angles are allowed. An equation can have several solutions in one full turn and more solutions when the domain covers multiple turns. The goal is to find every allowed solution and no others. This lesson focuses on first-degree equations and second-degree equations that can be solved by treating a repeated trigonometric ratio like a variable.

What you will learn

Use reference angles and quadrant signs to solve basic sine and cosine equations.
Find every solution in a stated domain.
Solve second-degree trigonometric equations by substitution and factoring.
Check solutions in the original equation and report exact values when possible.

1. Review: angles, ratios, and reference angles

Sine and cosine give coordinates on the unit circle, a circle with radius one centred at the origin. At an angle measured from the positive horizontal axis, the point on the circle has coordinates

(\cos x,\sin x)

. This connects sine and cosine values and signs to the angle’s position.

A reference angle is the acute angle between an angle’s terminal arm and the horizontal axis. Find the reference angle for the ratio’s value, then use the sign of sine or cosine to identify the matching quadrants. Sine is positive above the horizontal axis. Cosine is positive to the right of the vertical axis.

A degree measure and a radian measure describe the same angle in different units. A full turn is

360^\circ

or

2\pi

radians. Use the unit specified in the question. Unless stated otherwise, the examples in this lesson use degrees.

Read the domain before solving. The interval

0^\circ\leq x<360^\circ

includes

0^\circ

but excludes

360^\circ

. These angles point in the same direction, so this interval includes only one representation of that position.

(\cos x,\sin x)

Find a reference angle, then use quadrant signs.
Check whether each domain endpoint is included.

2. First-degree equations: solve, then apply the domain

A first-degree trigonometric equation has a trigonometric ratio to the first power, such as

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

or

2\cos x-1=0

. Use ordinary algebra to isolate the ratio. Then find every angle in the domain that gives that value.

For

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

on

0^\circ\leq x<360^\circ

, the reference angle is

30^\circ

. Sine is positive in Quadrants I and II, so the angles are

30^\circ

and

150^\circ

. Both are in the stated domain.

For cosine, use the same method but remember that cosine is positive in Quadrants I and IV and negative in Quadrants II and III. For tangent, use its reference angle and sign in each quadrant. Tangent is undefined where cosine is zero, so those angles cannot be solutions to an equation that requires a defined tangent value.

A domain may cover more than one turn. Continue listing angles that produce the required ratio until reaching the end of the domain. Do not stop at a familiar pair of answers without checking the interval.

\sin x=\frac{1}{2}\;\Longrightarrow\;x=30^\circ,150^\circ\quad(0^\circ\leq x<360^\circ)

Isolate the trigonometric ratio before finding angles.
Use the ratio’s sign to choose quadrants.
Keep only angles inside the stated domain.

3. Second-degree equations: substitute and factor

A second-degree trigonometric equation has a squared ratio, or a product of expressions involving the same ratio. For example,

2\sin^2 x-3\sin x+1=0

is quadratic in sine. Temporarily replace the repeated ratio with a variable.

Let

u=\sin x

. The equation becomes an ordinary quadratic in

u

. Solve it by factoring or another suitable algebra method. Then return to the trigonometric equation for each value of

u

. A value such as

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

can give two angles in one full turn, while

\sin x=1

gives one.

Sine and cosine values must be between

-1

and

1

, inclusive. If algebra gives a value outside this range, it cannot be the sine or cosine of an angle. Reject it. For each valid value, find all matching angles in the stated domain.

When both sine and cosine appear, first look for a direct algebraic rearrangement or an identity needed to solve the equation. Do not square both sides automatically. Squaring can create extra candidates, so check any candidates in the original equation.

The repeated ratio could be sine or cosine, depending on the equation. Choose a temporary variable that clearly represents that ratio.

Replace the repeated trigonometric ratio with a temporary variable.
Solve the quadratic, then solve a trigonometric equation for each root.
Check sine and cosine values against the range from $-1$ to $1$ .

4. A reliable routine and practice

Use a consistent routine. First, record the domain and angle unit. Next, isolate the ratio or substitute a variable if the equation is quadratic. Solve the resulting algebra carefully. Find angles for each valid ratio value, and keep only those inside the domain. Finally, substitute the angles into the original equation when possible.

Exact answers use familiar angle values or exact expressions, such as

150^\circ

or

\frac{\pi}{3}

. An approximate answer is a rounded decimal and should be identified as approximate. If a question allows a calculator and the ratio does not have a familiar exact value, follow the requested rounding.

For practice, solve

\cos x=-\frac{\sqrt{2}}{2}

on

0^\circ\leq x<360^\circ

. Then solve

2\cos^2 x-1=0

on the same domain. The first task tests signs and reference angles. The second requires solving for the cosine values before finding angles.

A useful sequence is to find valid ratio values first, then find the matching angles in the domain.

Keep algebraic candidates separate from final angle solutions.
Check the domain and the original equation before finishing.
State whether a numerical answer is exact or approximate.

Worked example

Example 1: A first-degree sine equation

Solve

2\sin x-\sqrt{3}=0

for

0^\circ\leq x<360^\circ

.

Reference angle for sine

The reference angle is $60^\circ$. Positive sine occurs in Quadrants I and II.

Isolate sine
Add $\sqrt{3}$ to both sides and divide by $2$ . These algebra steps preserve equality and give the sine value to match.
$\sin x=\frac{\sqrt{3}}{2}$
Find the reference angle
The positive sine value $\frac{\sqrt{3}}{2}$ has reference angle $60^\circ$ . Sine is positive in Quadrants I and II.
$x=60^\circ,120^\circ$
Apply the domain
Both angles are at least $0^\circ$ and less than $360^\circ$ , so both are allowed.

Answer:

x=60^\circ,120^\circ

Check: At either angle,

\sin x=\frac{\sqrt{3}}{2}

, so

2\sin x-\sqrt{3}=0

.

Worked example

Example 2: A quadratic in sine

Solve

2\sin^2 x-3\sin x+1=0

for

0^\circ\leq x<360^\circ

.

Angles for two sine values

The sine values are $\frac{1}{2}$ and $1$. Their angles in the stated domain are shown.

Substitute a variable
Let $u=\sin x$ . This turns the equation into a quadratic while keeping track of which ratio each root represents.
$2u^2-3u+1=0$
Factor and solve
Factor the quadratic and set each factor equal to zero. The two roots are the possible sine values. (2u-1)(u-1)=0\Longrightarrow u= $\frac{1}{2}$ or u=1
Find angles for each value
For $\sin x=\frac{1}{2}$ , the reference angle is $30^\circ$ and sine is positive in Quadrants I and II. For $\sin x=1$ , the angle is at the top of the unit circle.
$x=30^\circ,90^\circ,150^\circ$
Report the solutions
All three angles lie in the stated interval. List them in increasing order for a clear final answer.

Answer:

x=30^\circ,90^\circ,150^\circ

Check: At these angles, the sine values are

\frac{1}{2}

,

1

, and

\frac{1}{2}

. Each is a root of the factored quadratic.

Worked example

Example 3: A quadratic in cosine over two turns

Solve

2\cos^2 x-1=0

for

0^\circ\leq x<720^\circ

.

Cosine values and reference angle

The unit circle shows the four angles in each turn. The domain includes two turns but excludes $720^\circ$.

Solve for cosine values
Let $u=\cos x$ . Solving the quadratic gives two possible cosine values, and both are between $-1$ and $1$ .
$2u^2-1=0\Longrightarrow u=\pm\frac{\sqrt{2}}{2}$
Find angles in one turn
The reference angle for either value is $45^\circ$ . Positive cosine occurs in Quadrants I and IV; negative cosine occurs in Quadrants II and III.
$45^\circ,135^\circ,225^\circ,315^\circ$
Include the next turn
The domain extends to, but does not include, $720^\circ$ . Add one full turn, $360^\circ$ , to each solution from the first turn. Each resulting angle remains in the domain.
$405^\circ,495^\circ,585^\circ,675^\circ$

Answer:

x=45^\circ,135^\circ,225^\circ,315^\circ,405^\circ,495^\circ,585^\circ,675^\circ

Check: At every listed angle,

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

, so

2\cos^2 x-1=0

. The endpoint

720^\circ

is excluded.

Common mistakes and how to avoid them

Giving only one angle for a sine or cosine value.

Correction: Use the sign of the ratio to identify every matching quadrant, then check the domain.

Treating the roots of a quadratic in sine or cosine as angles.

Correction: The roots are ratio values. Solve a separate trigonometric equation for each root.

Including a domain endpoint that is excluded.

Correction: Read the inequality symbols carefully. For

0^\circ\leq x<360^\circ

, do not include

360^\circ

.

Keeping a ratio value outside the interval from

-1

to

1

.

Correction: Reject it because sine and cosine cannot take values outside that range.

Lesson summary

Isolate the trigonometric ratio in a first-degree equation.
For a second-degree equation, substitute a variable for the repeated ratio and solve the resulting quadratic.
Use reference angles and quadrant signs to find every matching angle.
Apply the stated domain and verify final solutions in the original equation.

Check your understanding

Question 1

Solve

\cos x=-\frac{\sqrt{2}}{2}

for

0^\circ\leq x<360^\circ

.

$x=45^\circ,315^\circ$
$x=135^\circ,225^\circ$
$x=135^\circ$
$x=45^\circ,135^\circ,225^\circ,315^\circ$

Show answer and explanation

x=135^\circ,225^\circ

The reference angle is

45^\circ

. Cosine is negative in Quadrants II and III, giving

135^\circ

and

225^\circ

.

Question 2

How many solutions does

\sin x=1

have on

0^\circ\leq x<360^\circ

?

None
One: $90^\circ$
Two: $90^\circ$ and $270^\circ$
One: $270^\circ$

Show answer and explanation

One:

90^\circ

Sine equals

1

only at the top point of the unit circle, which corresponds to

90^\circ

in this domain.

Question 3

For

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

, what does

\frac{1}{2}

represent?

An angle
A sine ratio value
The domain endpoint
A reference angle in degrees

Show answer and explanation

A sine ratio value

It is the value of the sine ratio. The angles in the stated domain still need to be found.

Key terms

Domain: The set or interval of input values allowed in a question.
Reference angle: The acute angle between an angle’s terminal arm and the horizontal axis.
Quadrant: One of the four regions formed by the horizontal and vertical axes.
Exact value: A value written precisely, such as $\frac{\sqrt{2}}{2}$ , rather than as a rounded decimal.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Alberta Grade 12 Mathematics (Math 30-1), expectation TR5. 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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TR5 · Solve first- and second-degree trigonometric equations with stated domains

Alberta Grade 12 Mathematics

TR5: Find every solution in the stated domain

What you will learn

1. Review: angles, ratios, and reference angles

2. First-degree equations: solve, then apply the domain

3. Second-degree equations: substitute and factor

4. A reliable routine and practice

Example 1: A first-degree sine equation

Example 2: A quadratic in sine

Example 3: A quadratic in cosine over two turns

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through Math 30-1

About this lesson and its review