Trigonometry
Grade 8-9Trigonometric Equations
Because the sine, cosine and tangent graphs repeat and fold back on themselves, most trig equations have more than one solution within a given range, and finding every one of them means using the symmetry of the graph, not just the calculator's first answer. This lesson covers solving basic trig equations, finding all solutions between 0° and 360°, solving equations with negative values, and equations that come from real-world models.
Written by Asad, Co-Founder of Teachably
Video walkthrough coming soon
The written lesson below covers everything you need in the meantime.
Solving a basic trig equation
For an equation like sin x° = k restricted to 0° to 90°, apply the inverse trig function directly to find x.
sin x° = 0.6 for 0 ≤ x ≤ 90°: x = sin⁻¹(0.6) = 36.9° (1 d.p.).
Finding all solutions between 0° and 360°
The calculator only gives one solution, but the graph's symmetry usually gives a second one within 0° to 360°. For sine, the second solution is 180° minus the first. For cosine, the second solution is 360° minus the first. For tangent, the second solution is 180° plus the first.
cos x° = 0.35 for 0 ≤ x ≤ 360°: the calculator gives x = cos⁻¹(0.35) = 69.5°, and the second solution is 360° - 69.5° = 290.5°.
Solving equations with a negative value
When the given value is negative, first find the positive reference angle by applying the inverse trig function to the positive version of the value, then use that reference angle to locate the solutions in the correct quadrants for the original sign.
tan x° = -1.4 for 0 ≤ x ≤ 360°: the reference angle is tan⁻¹(1.4) = 54.5°. Since tangent is negative in the second and fourth quadrants, the solutions are 180° - 54.5° = 125.5° and 360° - 54.5° = 305.5°.
Solving equations from a real-world model
When a trig equation is embedded in a real-world model, such as d = a + b cos x°, rearrange the equation first to isolate the trig term, then solve as normal.
The depth of water in a harbour is modelled by d = 6 + 3 cos x°. To find x when the depth is first 8 m: 3 cos x° = 8 - 6 = 2, so cos x° = 2 ÷ 3 = 0.667, giving x = cos⁻¹(0.667) = 48.2° (1 d.p.).
Worked Examples
Three exam-style questions, fully solved.
Solve sin x° = 0.6 for 0 ≤ x ≤ 90°. Give your answer correct to 1 decimal place.
Easy- 1.Apply the inverse sine function directly: x = sin⁻¹(0.6)
Answer: 36.9°
Solve tan x° = -1.4 for 0 ≤ x ≤ 360°. Give your answers correct to 1 decimal place.
Medium- 1.Find the reference angle using the positive value: tan⁻¹(1.4) = 54.5°
- 2.Tangent is negative in the second and fourth quadrants, so use 180° - 54.5° and 360° - 54.5°
Answer: 125.5° or 305.5°
The depth of water in a harbour, d metres, is modelled by d = 6 + 3 cos x°, where x° represents the time since high tide. Find the value of x (0 ≤ x ≤ 360°) when the depth is first 8 m. Give your answer correct to 1 decimal place.
Hard- 1.Rearrange to isolate the trig term: 3 cos x° = 8 - 6 = 2
- 2.Divide to isolate cos x°: cos x° = 2 ÷ 3 = 0.667
- 3.Apply the inverse cosine function: x = cos⁻¹(0.667)
Answer: 48.2°
Avoid These
The most common mistakes students make.
Only giving one solution within 0° to 360° when the equation actually has two valid solutions.
Using the wrong rule to find the second solution, for example using 180° - x for cosine instead of 360° - x.
Applying the inverse trig function directly to a negative value, instead of finding the positive reference angle first and then locating the correct quadrants.
Forgetting to isolate the trig function before applying the inverse function, for example trying sin⁻¹ on "2 sin x°" directly instead of on "sin x°" alone.
In a real-world modelling context, forgetting to rearrange the model equation fully, such as not subtracting the baseline value before dividing by the amplitude.
FAQ
Questions parents and students ask.
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