PreCalc 2 · Analytic Trigonometry
Solve on a given interval. When the domain is all real numbers, write every solution by adding the period.
Restricted domain
A problem that gives an interval, such as [0, 2π), wants only the solutions inside it.
Isolate one trig function. Find the reference angle. Then use the quadrants where that function has the right sign.
All real solutions
When the domain is all real numbers, add the period.
Sine and cosine repeat every 2π.
Tangent repeats every π.
k stands for any integer.
Algebra first
A quadratic in sin θ or cos θ factors like any quadratic.
Set each factor equal to 0.
Then solve each trig equation.
A double angle
For sin 2θ = a number, solve for 2θ first.
If θ is in [0, 2π), then 2θ is in [0, 4π).
Divide by 2 at the end.
One trig function. Isolate it, then read the solutions.
cos θ = 2
Cosine is positive in Quadrant I and Quadrant IV.
The reference angle is π4.
Quadrant I: θ = π4
Quadrant IV: θ = 2π − π4
8π4 − π4
7π4
Answer: θ = π4, or θ = 7π4
2 sin θ − 1 = 0
2 sin θ = 1
sin θ = 12
Sine is positive in Quadrant I and Quadrant II.
The reference angle is π6.
Quadrant I: θ = π6
Quadrant II: θ = π − π6
6π6 − π6
5π6
Answer: θ = π6, or θ = 5π6
cos θ + 1 = 0
cos θ = −1
θ = π
Answer: θ = π
tan θ =
Tangent is positive in Quadrant I and Quadrant III.
The reference angle is π3.
Quadrant I: θ = π3
Answer: θ = π3
sin θ = −2
Sine is negative in Quadrant III and Quadrant IV.
The reference angle is π4.
Quadrant III: θ = π + π4
4π4 + π4
5π4
Quadrant IV: θ = 2π − π4
8π4 − π4
7π4
Answer: θ = 5π4, or θ = 7π4
cos θ = 0
Cosine is 0 on the y-axis.
θ = π2
θ = 3π2
Answer: θ = π2, or θ = 3π2
sin θ = 2
Sine is positive in Quadrant I and Quadrant II.
The reference angle is π3.
Add the sine period, 2π, to each solution.
k is any integer.
θ = π3 + 2πk
θ = 2π3 + 2πk
Answer: θ = π3 + 2πk, or θ = 2π3 + 2πk, where k is any integer.
cos θ = 12
Cosine is positive in Quadrant I and Quadrant IV.
The reference angle is π3.
Add the cosine period, 2π, to each solution.
k is any integer.
θ = π3 + 2πk
θ = −π3 + 2πk
Answer: θ = π3 + 2πk, or θ = −π3 + 2πk, where k is any integer.
tan θ = −1
Tangent is negative in Quadrant II and Quadrant IV.
The reference angle is π4.
One solution in the arctan range is −π4.
Tangent has period π.
k is any integer.
θ = −π4 + πk
Answer: θ = −π4 + πk, where k is any integer.
sin θ = −1
sin θ = −1 at 3π2, and then every full turn.
Add the sine period, 2π.
k is any integer.
θ = −π2 + 2πk
Answer: θ = −π2 + 2πk, where k is any integer.
cos θ = −1
cos θ = −1 at π, and then every full turn.
Add the cosine period, 2π.
k is any integer.
θ = π + 2πk
Answer: θ = π + 2πk, where k is any integer.
Factor, or use a double angle, before you list the solutions.
2 sin² θ − 1 = 0
2 sin² θ = 1
sin² θ = 12
sin θ = 2 or sin θ = −2
The reference angle is π4.
Quadrant I: θ = π4
Quadrant II: θ = π − π4
4π4 − π4
3π4
The reference angle is π4.
Quadrant III: θ = π + π4
4π4 + π4
5π4
Quadrant IV: θ = 2π − π4
8π4 − π4
7π4
Answer: θ = π4, θ = 3π4, θ = 5π4, or θ = 7π4
2 sin² θ − sin θ − 1 = 0
Let u = sin θ.
2u² − u − 1 = 0
(2u + 1)(u − 1) = 0
2u + 1 = 0 or u − 1 = 0
sin θ = −12 or sin θ = 1
The reference angle is π6.
Quadrant III: θ = π + π6
6π6 + π6
7π6
Quadrant IV: θ = 2π − π6
12π6 − π6
11π6
The reference angle is π2.
Quadrant I: θ = π2
Answer: θ = π2, θ = 7π6, or θ = 11π6
sin 2θ = 12
Solve for 2θ, then divide by 2. 2θ runs through [0, 4π) when θ is in [0, 2π).
2θ = π6
θ = π12
2θ = 5π6
θ = 5π12
2θ = 13π6
θ = 13π12
2θ = 17π6
θ = 17π12
Answer: θ = π12, θ = 5π12, θ = 13π12, or θ = 17π12
cos 2θ = 0
Solve for 2θ, then divide by 2. 2θ runs through [0, 4π) when θ is in [0, 2π).
2θ = π2
θ = π4
2θ = 3π2
θ = 3π4
2θ = 5π2
θ = 5π4
2θ = 7π2
θ = 7π4
Answer: θ = π4, θ = 3π4, θ = 5π4, or θ = 7π4
tan² θ = 1
tan θ = 1 or tan θ = −1
The reference angle is π4.
Quadrant I: θ = π4
Quadrant III: θ = π + π4
4π4 + π4
5π4
The reference angle is π4.
Quadrant II: θ = π − π4
4π4 − π4
3π4
Quadrant IV: θ = 2π − π4
8π4 − π4
7π4
Answer: θ = π4, θ = 3π4, θ = 5π4, or θ = 7π4
2 cos² θ + cos θ − 1 = 0
Let u = cos θ.
2u² + u − 1 = 0
(2u − 1)(u + 1) = 0
2u − 1 = 0 or u + 1 = 0
cos θ = 12 or cos θ = −1
The reference angle is π3.
Quadrant I: θ = π3
Quadrant IV: θ = 2π − π3
6π3 − π3
5π3
θ = π
Answer: θ = π3, θ = π, or θ = 5π3
2 cos θ + 1 = 0
2 cos θ = −1
cos θ = −12
Cosine is negative in Quadrant II and Quadrant III.
The reference angle is π3.
Add the cosine period, 2π.
k is any integer.
θ = 2π3 + 2πk
θ = −2π3 + 2πk
Answer: θ = 2π3 + 2πk, or θ = −2π3 + 2πk, where k is any integer.
sin² θ = 12
sin θ = 2 or sin θ = −2
The reference angle is π4.
These four quadrants repeat every π.
k is any integer.
θ = π4 + πk
θ = −π4 + πk
Answer: θ = π4 + πk, or θ = −π4 + πk, where k is any integer.
tan θ =
Tangent is positive in Quadrant I and Quadrant III.
The reference angle is π3.
Tangent has period π, so Quadrant III is the next period of the Quadrant I solution.
k is any integer.
θ = π3 + πk
Answer: θ = π3 + πk, where k is any integer.
sin 2θ = −1
2θ = 3π2 + 2πk
Divide by 2.
θ = 3π4 + πk, where k is any integer
k is any integer.
Answer: θ = 3π4 + πk, where k is any integer.
cos 2θ = 12
Write every solution for 2θ, then divide by 2.
2θ = π3 + 2πk
θ = π6 + πk
2θ = −π3 + 2πk
θ = −π6 + πk
k is any integer.
Answer: θ = π6 + πk, or θ = −π6 + πk, where k is any integer.
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