Grade by Grade
B. Trigonometric Functions · Trigonometric Identities and Equations

Linear Equations with Multiple and Shifted Angles

B3.4About 70 minutes
Learning goals
  • I can solve equations such as 2 cos 2x = 1, tan 3x = 1 and sin(x − π/4) = √3/2 for 0 ≤ x ≤ 2π.
  • I can find the new interval for θ = kx + c when 0 ≤ x ≤ 2π.
  • I can predict the number of solutions from the period of the function.
  • I can convert solutions for θ back into solutions for x and check them.

Words to know

multiple angle
An angle such as 2x, 3x or x/2, a constant multiple of the variable.
substitution
Replacing an expression with a single letter, such as θ = 2x, to turn a new problem into a familiar one.
period
The horizontal length of one cycle. y = sin kx and y = cos kx have period 2π/k; y = tan kx has period π/k (k > 0).

More waves, more solutions

In the unit Radian Measure and Trigonometric Functions you saw that y = cos 2x has period π: it fits two full waves between 0 and 2π. So a horizontal line such as y = 1/2 crosses it four times instead of two. Equations with 2x, 3x or (x − π/4) inside a ratio need a little extra care so that you find every solution and no extras.

xπ/2π3π/22π11/2−1π/65π/67π/611π/6y = cos 2x
y = cos 2x completes two cycles between 0 and 2π, so the line y = 1/2 crosses it four times: x = π/6, 5π/6, 7π/6 and 11π/6.

The substitution method

  1. Substitute θ for the angle inside the ratio, for example θ = 2x.
  2. Change the interval. If 0 ≤ x ≤ 2π, then 0 ≤ 2x ≤ 4π, so 0 ≤ θ ≤ 4π. For θ = x − π/4, the interval becomes −π/4 ≤ θ ≤ 7π/4.
  3. Solve for θ as in the last lesson, then keep adding 2π (π for tangent) to each answer while it stays in the θ-interval. Also check whether subtracting 2π gives a value still in the interval.
  4. Convert back to x (for θ = 2x, divide by 2) and check that every x is in 0 ≤ x ≤ 2π.
θx0π2π3π4π0π/2π3π/22ππ/3π/65π/35π/67π/37π/611π/311π/6x = θ ÷ 2
Solving cos θ = 1/2 on 0 ≤ θ ≤ 4π gives four values. Dividing each by 2 maps them into 0 ≤ x ≤ 2π.
Worked example

Example 1: A double angle

Solve 2 cos 2x = 1 for 0 ≤ x ≤ 2π.

  1. cos 2x = 1/2. Let θ = 2x, with 0 ≤ θ ≤ 4π.
  2. cos θ = 1/2: α = π/3, cosine positive in quadrants I and IV: θ = π/3, 5π/3.
  3. Add 2π: θ = 7π/3, 11π/3 (both ≤ 4π). Adding 2π again goes past 4π.
  4. Divide by 2: x = π/6, 5π/6, 7π/6, 11π/6.
  5. Check one: cos(2 · 7π/6) = cos(7π/3) = cos(π/3) = 1/2 ✓.

x = π/6, 5π/6, 7π/6, 11π/6

Worked example

Example 2: An axis value

Solve sin 2x = −1 for 0 ≤ x ≤ 2π.

  1. θ = 2x, 0 ≤ θ ≤ 4π. sin θ = −1 at θ = 3π/2, and again at 3π/2 + 2π = 7π/2.
  2. x = θ/2 = 3π/4, 7π/4.

x = 3π/4, 7π/4

Worked example

Example 3: Tangent of a triple angle

Solve tan 3x = 1 for 0 ≤ x ≤ 2π.

  1. θ = 3x, 0 ≤ θ ≤ 6π. tan θ = 1 at θ = π/4, and tangent repeats every π.
  2. θ = π/4, 5π/4, 9π/4, 13π/4, 17π/4, 21π/4. (The next, 25π/4, is more than 6π = 24π/4.)
  3. Divide by 3: x = π/12, 5π/12, 9π/12, 13π/12, 17π/12, 21π/12.

x = π/12, 5π/12, 3π/4, 13π/12, 17π/12, 7π/4 (six solutions)

Worked example

Example 4: A shifted angle

Solve sin(x − π/4) = √3/2 for 0 ≤ x ≤ 2π.

  1. θ = x − π/4. Since 0 ≤ x ≤ 2π, −π/4 ≤ θ ≤ 7π/4.
  2. sin θ = √3/2: θ = π/3 or 2π/3. Both are in the interval. (Adding or subtracting 2π leaves it.)
  3. x = θ + π/4: π/3 + π/4 = 7π/12 and 2π/3 + π/4 = 11π/12.

x = 7π/12, 11π/12

Worked example

Example 5: Calculator values

Solve sin 2x = 0.6 for 0 ≤ x ≤ 2π, to two decimal places.

  1. θ = 2x, 0 ≤ θ ≤ 4π. α = sin⁻¹(0.6) ≈ 0.6435.
  2. Quadrants I and II: θ ≈ 0.6435 and π − 0.6435 ≈ 2.4981.
  3. Add 2π: θ ≈ 6.9267 and 8.7813 (both less than 4π ≈ 12.566).
  4. Divide by 2: x ≈ 0.3218, 1.2490, 3.4633, 4.3906.

x ≈ 0.32, 1.25, 3.46, 4.39

Worked example

Example 6: A half angle gives fewer solutions

Solve cos(x/2) = −1/2 for 0 ≤ x ≤ 2π.

  1. θ = x/2, so 0 ≤ θ ≤ π. Only half a turn!
  2. cos θ = −1/2 in quadrants II and III: 2π/3 and 4π/3. Only 2π/3 is in [0, π].
  3. x = 2θ = 4π/3.

x = 4π/3 (one solution)

Count before you solve

For y = sin kx or y = cos kx with k a positive whole number, there are k full cycles in 0 ≤ x ≤ 2π. A horizontal line strictly between −1 and 1 crosses each cycle twice, so expect 2k solutions. (Values ±1 and 0 can change the count, and endpoints may add one.) If your list is too short, you forgot to add 2π; if it is too long, you went past the interval.

Just for fun

Doubling the angle is like playing a song at double speed: twice as many beats fit into the same time, and twice as many solutions fit between 0 and 2π.

Hands-on activity

Count the crossings

You need: graph paper, a pencil, a ruler and a calculator in radian mode.

Step 1. Label an x-axis from 0 to 6.3 (about 2π) and a y-axis from −1 to 1.

Step 2. Make a table of y = sin 2x for x = 0, 0.2, 0.4, ..., 6.2 and plot the points. Join them with a smooth curve.

Step 3. Draw the horizontal line y = 0.6. Mark each crossing and read its x-value as accurately as you can.

Step 4. Compare with Example 5.

What to look for: two full waves and four crossings, near x ≈ 0.3, 1.2, 3.5 and 4.4. Your reading will be off by a little; the algebra gives the precise values 0.32, 1.25, 3.46 and 4.39.

Your turn

1

Solve 2 sin 2x = √3 for 0 ≤ x ≤ 2π.

2

Solve cos 3x = 0 for 0 ≤ x ≤ 2π.

3

Solve tan 2x = −1 for 0 ≤ x ≤ 2π.

4

Solve cos(x + π/6) = 1/2 for 0 ≤ x ≤ 2π.

5

Solve sin(x/2) = 1/2 for 0 ≤ x ≤ 2π.

6

Solve 2 cos 2x + 1 = 0 for 0 ≤ x ≤ 2π.

More practice: warm-up

1

If 0 ≤ x ≤ 2π, what is the interval for 3x?

2

If 0 ≤ x ≤ 2π, what is the interval for x − π/3?

3

What is the period of y = sin 4x?

4

Solve sin 2x = 1 for 0 ≤ x ≤ 2π.

5

How many solutions does cos 2x = 0.3 have for 0 ≤ x ≤ 2π?

More practice: on your own

1

Solve 3 tan 2x = √3 for 0 ≤ x ≤ 2π.

2

Solve cos 2x = 0.25 for 0 ≤ x ≤ 2π, to two decimal places.

3

Solve sin(2x − π/3) = 0 for 0 ≤ x ≤ 2π.

4

Solve 2 sin(x + π/4) = −√2 for 0 ≤ x ≤ 2π.

5

Without solving, explain why sin 3x = 0.5 has six solutions for 0 ≤ x ≤ 2π.

Challenge

1

Solve tan(x/2) = √3 for 0 ≤ x ≤ 2π.

2

Solve 2 cos(3x − π) + 1 = 0 for 0 ≤ x ≤ 2π.

3

For which positive integer k does sin kx = 1 have exactly 3 solutions for 0 ≤ x ≤ 2π?

Check your understanding

Quiz

1.

If 0 ≤ x ≤ 2π, what interval should you use for θ = 2x?

2.

What are the solutions of sin 2x = 1 for 0 ≤ x ≤ 2π?

3.

How many solutions does tan 3x = 1 have for 0 ≤ x ≤ 2π?

4.

What is the solution set of cos(x/2) = −1/2 for 0 ≤ x ≤ 2π?

5.

What are the solutions of sin(x − π/4) = √3/2 for 0 ≤ x ≤ 2π?

6.

What is the period of y = sin 3x?

For parents and teachers

Big idea (B3.4): when the angle inside a ratio is 2x, 3x, x/2 or x − c, substitute θ for it, adjust the interval (0 ≤ x ≤ 2π becomes 0 ≤ 2x ≤ 4π, and so on), solve for θ, and convert back. More cycles mean more solutions.

How to help: Before your teen solves, ask: "How many solutions do you expect?" (Count the cycles.) Afterwards, have them check one answer by substituting it back in on a calculator in radian mode.

Common difficulties:

  • Solving for θ only on [0, 2π] and so missing half the answers for 2x.
  • Forgetting to divide (or shift) at the end.
  • For shifted angles, forgetting that the θ-interval starts at a negative value.
  • For x/2, keeping answers that are past 2π after doubling.