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.
What you need to know
- A trig equation usually has more than one solution in a given interval.
- Find the first solution with the inverse, then use the graph or symmetry for the rest.
- sin is positive in the first and second quadrants, cos in the first and fourth, tan in the first and third.
- Check that every answer lies in the range the question asks for.
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°
Practice
6 questions, marked instantly
Type an answer and check it. The worked solution appears once you have had a go. No login needed, and your progress saves in this browser.
Practice
Now try these yourself.
Type your answer and check it. The worked solution appears once you have had a go.
Solve sin x° = 0.5 for 0° ≤ x ≤ 180°.
Solve cos x° = 0 for 0° ≤ x ≤ 360°.
Solve tan x° = 1 for 0° ≤ x ≤ 360°.
Solve sin x° = 1 for 0° ≤ x ≤ 360°.
One solution of cos x° = 0.6 is x = 53.1°. Work out the other solution for 0° ≤ x ≤ 360°.
One solution of sin x° = 0.4 is x = 23.6°. Work out the other solution for 0° ≤ x ≤ 180°.
Avoid these
The mistakes students make most often
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 students and parents ask
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