Find All Solutions in a Period

A basic trigonometric equation is all that is required to solve any complex expression that starts with many trig terms. Our job as trig solvers is to discover the combination of identities and algebraic manipulations that will result in a basic trigonometric equation.

Solving a Basic Trigonometric Equation

Example 01: An equation with multiple solutions

Solve the equation $\sin(x)=1$.

Solution 01:

We will break the method of this solution into two parts.

Find the solutions in one period.

Recall that $\sin$ is a periodic function, with a period of $2\pi$. That means after $2\pi$ the function repeats itself. So our first step in solving when $\sin(x)=1$ is to find all possible solutions in the range of that periods, $0\leq x\leq 2\pi$.

We can use a unit circle or a graph to locate the solutions.

The graph on the left is a single period of a sine wave. We can find when $\sin(x)=1$ by drawing a line across at $y=1$ and finding the points where it intersects with the sine wave. This occurs at the point $x=\pi /2$ The unit circle is on the right, and we can see that $\sin(x)=1$ when the circle is at $(0,1)$, indicating a solution of $\pi /2$.

Find all Solutions

Because the sine function is periodic and repeats ever $2\pi$, we can find every solution to $\sin(x)=1$ by adding $2\pi$ any amount of times to the solution above.

$$ \sin(x)=1 \quad \Rightarrow \quad x=\frac{\pi}{2} \quad \text{so} \quad x = \frac{\pi}{2} + 2\pi n$$

Where $n$ is any integer. You can try it for yourself. Try putting $\sin\left((\pi /2) + 2\pi (5) \right) $ or $\sin\left((\pi /2) + 2\pi (507) \right) $into your calculator – it will still be equal to 1.

Example 02: Solving a Basic Trigonometric Equation by Factoring

Solve the equation $2\cos(x)\sin(x) + \sin(x) = 0$

Solution 02:

We can factor the left-hand-side of the equation:

$$\begin{align} 2\cos(x)\sin(x)+\sin(x) & = 0 &&\text{Given equation} \\[1em] \sin(x)\left(2\cos(x) + 1\right) &= 0 &&\text{Factor} \\[1em] \sin(x) = 0 \quad \text{or} \quad 2\cos(x) + 1 &= 0 &&\text{Set each factor equal to zero}\\[1em] \cos(x)&=-\frac{1}{2} \end{align} $$

We have two equations that can be set equal to zero, so the solution to this problem is any angle that makes $\sin(x) = 0$ or makes $\cos(x)=-1/2$.

We start with $\cos(x)=-1/2$

As we can see in the graph, and in the unit circle, the solutions for $\cos(x)=-1/2$ in the first period is when $x=2\pi/3$ and when $x=4\pi/3$.

Now we look at when $\sin(x)=0$

As we can see from the graph and the unit circle, the solutions for $\sin(x)=0$ in the first period is when $x=0$ and $x=\pi$.

Now we have all 4 solutions to the original basic trig equation, we just have to add the $2\pi n$ to cover all possible answers.

$$\begin{align}&x=\pi n && x=\frac{2\pi}{3} + 2\pi n && x = \frac{4\pi}{3} + 2\pi n\end{align}$$

Notice that we can combine the two solutions for $\sin(x)=0$ because they include all multiples of $\pi$. So it is enough to write $x=\pi n$ because $n$ can be 0, or 1, or 2. It will work for all integers.

About the Author

James Lowman is an applied mathematician currently working on a Ph.D. in the field of computational fluid dynamics at the University of Waterloo. He is a co-founder of the online math and science tutoring company Waterloo Standard.

If you are looking for assistance with math, book a session with James.

Find All Solutions in a Period

Source: https://waterloostandard.com/post/basic-trigonometric-equations/

Statton Inathind