5.10 Solving trigonometric equations in an interval — Learn

Prerequisites: the graphs of the three functions; the principal values of the inverse functions; the CAST quadrants; solving a quadratic by factorising.

By the end of this section you can: solve an equation of the form $\sin\theta = k$, $\cos\theta = k$ or $\tan\theta = k$ in a given interval; find all solutions without losing any; solve an equation in $\sin 2\theta$ or $\cos 3\theta$; solve a quadratic in a trigonometric function.

1. One value, many angles

A trigonometric equation usually has more than one solution in an interval, because the graphs repeat. The calculator gives one angle, called the principal value. The graphs then give the others.

For $0 \le \theta < 2\pi$:

• $\sin\theta = k$ has principal value $\arcsin k$ in the first or fourth quadrant. The second solution is $\pi - \arcsin k$ when $k > 0$, or $\pi + \arcsin k$ when $k < 0$.
• $\cos\theta = k$ has principal value $\arccos k$. The second solution is $2\pi - \arccos k$.
• $\tan\theta = k$ has period $\pi$, so the second solution in the interval is $\arctan k + \pi$.

A quick sketch of the curve between $0$ and $2\pi$ shows how many solutions to expect. Sine and cosine each give two solutions for a value strictly between $-1$ and $1$. Tangent gives two solutions over a full turn.

2. Worked examples

Example 1. Solve $\sin\theta = \dfrac{1}{2}$ for $0 \le \theta < 2\pi$.

One angle with sine $\dfrac{1}{2}$ is $\dfrac{\pi}{6}$. The other is

The solutions are $\theta = \dfrac{\pi}{6}$ and $\theta = \dfrac{5\pi}{6}$.

Example 2. Solve $\cos\theta = -\dfrac{\sqrt{3}}{2}$ for $0 \le \theta < 2\pi$.

The related acute angle is $\dfrac{\pi}{6}$, and cosine is negative in the second and third quadrants. So

Example 3. Solve $\tan\theta = 1$ for $0 \le \theta < 2\pi$.

The principal value is $\dfrac{\pi}{4}$. Tangent repeats every $\pi$, so the second solution is $\dfrac{\pi}{4} + \pi = \dfrac{5\pi}{4}$.

3. A multiple angle

When the equation involves $2\theta$, solve first for the whole angle $2\theta$, then divide at the end. Change the interval for $2\theta$ at the same time.

Example 4. Solve $\sin 2\theta = \dfrac{1}{2}$ for $0 \le \theta < 2\pi$.

Let $x = 2\theta$. Since $0 \le \theta < 2\pi$, the interval for $x$ is $0 \le x < 4\pi$. The equation $\sin x = \dfrac{1}{2}$ has solutions $\dfrac{\pi}{6}$ and $\dfrac{5\pi}{6}$ in the first turn, and two more in the second turn:

Divide each by $2$:

4. A quadratic in a trigonometric function

Some equations are quadratics once you treat $\sin\theta$ as the unknown.

Example 5. Solve $2\sin^2\theta - \sin\theta - 1 = 0$ for $0 \le \theta < 2\pi$.

Factorise:

So $\sin\theta = -\dfrac{1}{2}$ or $\sin\theta = 1$.

For $\sin\theta = -\dfrac{1}{2}$, the related acute angle is $\dfrac{\pi}{6}$, and sine is negative in the third and fourth quadrants:

For $\sin\theta = 1$, the only solution in the interval is $\theta = \dfrac{\pi}{2}$.

The solutions are $\dfrac{\pi}{2}$, $\dfrac{7\pi}{6}$ and $\dfrac{11\pi}{6}$.

5. Try it: counting the solutions

Predict first. For $\sin\theta = k$ with $k$ between $-1$ and $1$, how many solutions lie in the interval $0 \le \theta < 2\pi$? How many does $\sin 2\theta = k$ have in the same interval? Write your prediction down before you touch the controls.

Test. Choose a function and move the slider for $k$. Each dot marks a crossing of the horizontal line with the curve, and the panel lists the angles.

Explain. Set the function to $\sin\theta$ and then to $\sin 2\theta$ with the same $k$. Why does the second equation have more solutions in the same interval? Write two sentences linking the number of solutions to the number of turns of the curve.

Watch out. Three mistakes are common when solving these equations.

Stopping at the principal value. The calculator gives one angle. In a full turn, sine and cosine usually give a second angle, and tangent gives a second. Sketch the curve or use the CAST diagram to find every solution in the interval.

Dividing by a trigonometric function. From an equation such as $\sin\theta\cos\theta = \sin\theta$, dividing by $\sin\theta$ removes the solutions where $\sin\theta = 0$. Factorise instead: $\sin\theta(\cos\theta - 1) = 0$.

Forgetting to change the interval for a multiple angle. If $0 \le \theta < 2\pi$ and the equation uses $2\theta$, then $2\theta$ runs from $0$ to $4\pi$, which is two full turns. Missing the second turn loses half the solutions. Re-check: count the solutions your sketch predicts, and confirm that your list has that many.

You should now be able to

• find the principal value of a trigonometric equation;
• find every solution in a stated interval using the graph or the CAST diagram;
• solve an equation in a multiple angle by changing the interval;
• solve a quadratic in sine or cosine;
• factorise rather than divide when a common trigonometric factor appears;
• count the expected number of solutions before you start.