The Trig Equation Solver
Topics 9.9 -- 9.12: Solving equations in sine and cosine
Step 1 of 8
How many solutions does sin x = 0.5 have?
The sine curve is drawn with a horizontal line at \(y = 0.5\). Click every intersection you can see.
Intersections found: 0
Step 2 of 8
Why infinitely many?
Why does \(\sin x = 0.5\) have infinitely many solutions? Pick what you think.
Step 3 of 8
Two solutions per lap: the unit circle
On the unit circle, \(\sin\theta = 0.5\) means the \(y\)-coordinate is 0.5. A horizontal line at \(y = 0.5\) crosses the circle at exactly two points. Click on both points.
Points found: 0 / 2
Step 4 of 8
Write every solution: the general solution
The two base solutions are \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\). Since sine repeats every \(2\pi\), the general solution is:
Try different values of \(n\) and confirm each solution appears on the graph.
Step 5 of 8
Filtering to one interval
Most problems ask for solutions in \([0, 2\pi)\). That means: give me one lap. From the general solution, which values land in \([0, 2\pi)\)?
Step 6 of 8
Quadratic form: substitution
Consider the equation \(2\sin^2 x - \sin x - 1 = 0\). Look at the structure. What if you called \(\sin x\) by the letter \(u\)?
That is a quadratic you know how to factor. What does it factor to?
Step 7 of 8
Solve each branch on the circle
The factoring gives two branches: \(\sin x = -\frac{1}{2}\) and \(\sin x = 1\). How many total solutions in \([0, 2\pi)\)?
Step 8 of 8
Identity required: two trig functions become one
The equation \(2\cos^2 x + \sin x = 1\) has two different trig functions. You cannot factor it as is. What identity would you use to rewrite it in terms of sine only?
Explore Trig Equations
Solutions in \([0, 2\pi)\)
General Solution
Select an equation above to see its graph, unit circle representation, and complete solution set.
Try This
Part A: Solve \(2\cos x + 1 = 0\) on \([0, 2\pi)\).
Part B: Now write the general solution for \(2\cos x + 1 = 0\).
Part C: How many solutions does \(2\cos x + 1 = 0\) have on \([0, 10\pi)\)?
The harbor. Water depth is \(d(t) = 4\sin(\pi t / 6) + 8\) feet, where \(t\) is hours after midnight. Your boat needs 10 feet of clearance. During what hours of a 24-hour day can you leave?
First: at what value of \(d\) can you leave?
Part B: Solving \(\sin(\pi t/6) \ge \frac{1}{2}\), which intervals of the first 24 hours have enough depth? Give the boundary times.
Challenge. Solve \(2\sin^2 x - 3\cos x - 3 = 0\) on \([0, 2\pi)\). You will need an identity first.
Step 1: Which identity converts this to a single trig function?
Step 2: After substituting and simplifying, you get \(-2\cos^2 x - 3\cos x - 1 = 0\), or equivalently \(2\cos^2 x + 3\cos x + 1 = 0\). This factors to \((2\cos x + 1)(\cos x + 1) = 0\). What are the solutions on \([0, 2\pi)\)?
Step 3: Suppose instead the factoring gave you a branch \(\cos x = -2\). Why does that branch have no solutions?
- Explain why trig equations have infinitely many solutions and express the general solution using \(+ 2\pi n\).
- Find all solutions in a given interval, such as \([0, 2\pi)\), by identifying the base angles on the unit circle.
- Solve trig equations in quadratic form by substituting \(u = \sin x\) or \(u = \cos x\) and factoring.
- Use Pythagorean identities to convert an equation with two trig functions into one that can be factored.
- Recognize when a branch has no solution because the value is outside \([-1, 1]\).
Quick Check
How many solutions does \(\sin x = -1\) have on \([0, 2\pi)\)?
Instructor Notes
Teaching Notes
The critical shift for students is from "find the answer" to "find all the answers." Most have spent years in algebra where each equation has a single solution (or at most two for quadratics). The idea that an equation can have infinitely many solutions, organized by a repeating pattern, is genuinely new.
The unit circle is the bridge. When students can see that \(\sin\theta = 0.5\) means "where is \(y = 0.5\) on the circle," they naturally see two points per lap. The general solution then becomes obvious: add another lap.
Quadratic form is the biggest algebraic payoff in this section. Students who struggled with factoring in algebra often find it clicks here because the substitution step makes the structure visible. Emphasize that they already know how to factor -- the only new part is the substitution.
Common Student Errors
- One-answer syndrome: The calculator returns one value for \(\arcsin(0.5)\), and students stop. They need to learn that the calculator gives a reference angle, not the complete answer.
- Dividing by a trig function: Given \(\sin x \cos x = \sin x\), many divide both sides by \(\sin x\), silently losing every solution where \(\sin x = 0\). Emphasize: move everything to one side and factor instead.
- Forgetting the interval: Students write all solutions when the problem asks for \([0, 2\pi)\), or give only two when the problem asks for the general solution.
- Quadrant errors: When \(\sin x = -0.5\), students often place solutions in the wrong quadrants. The unit circle visualization helps: negative sine means the bottom half.
Discussion Questions
- Why does your calculator only give you one answer for \(\sin^{-1}(0.5)\) when we just showed there are infinitely many? What is the calculator actually returning?
- If someone solves \(\sin x \cos x = 0\) by dividing both sides by \(\cos x\), what solutions do they lose? How would you solve it instead?
- The general solution uses \(+ 2\pi n\). Why \(2\pi\) and not \(\pi\)? Can you think of an equation where the solutions repeat every \(\pi\) instead?
Exam Connection
Exam questions typically ask for solutions in \([0, 2\pi)\) or for the general solution. The most common exam format: (1) a basic equation like \(2\sin x - 1 = 0\), (2) a quadratic-form equation requiring substitution and factoring, and (3) an equation requiring an identity first. The stretch and challenge problems in Apply mirror these three tiers.