20.1: Basic trigonometric equations

In this section we solve equations such as:

$$\sin(x)=0.5 \nonumber$$

We know that $x=\dfrac{\pi}{6}$ solves this equation. However, there are also other solutions such as $x=\dfrac{5\pi}{6}$ or $x=\dfrac{13\pi}{6}$ that can easily be checked with the calculator. Below, we will study how to find all solutions of equations of the form

$$\sin(x)=c, \quad \cos(x)=c, \quad\text{ and }\quad \tan(x)=c \nonumber$$

We start with equations involving the tangent.

The equation $\tan(x)=c$

Example $\PageIndex{1}$

Solve for $x$: $\tan(x)=\sqrt{3}$

Solution

There is an obvious solution given by $x=\tan^{-1}(\sqrt{3})=\dfrac{\pi}{3}$, as we studied in the last section. However, we can look for all solutions of $\tan(x)=\sqrt{3}$ by studying the graph of the tangent function, that is, by finding all points where the graph of the $y=\tan(x)$ intersects with the horizontal line $y=\sqrt{3}$. Since the function $y=\tan(x)$ is periodic with period $\pi$, we see that the other solutions of $\tan(x)=\sqrt{3}$ besides $x=\dfrac{\pi}{3}$ are

$$\dfrac{\pi}{3}+\pi, \quad \dfrac{\pi}{3}+2\pi, \quad \dfrac{\pi}{3}+3\pi, \dots, \quad\text{ and }\quad \dfrac{\pi}{3}-\pi, \quad \dfrac{\pi}{3}-2\pi, \quad \dfrac{\pi}{3}-3\pi, \dots \nonumber$$

In general, we write the solution as

$$x=\dfrac{\pi}{3}+n\cdot \pi, \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

The graph also shows that these are indeed all solutions of $\tan(x)=\sqrt{3}$.

Observation: Solve $\tan(x)=c$

By the same argument we also get the general solution of $\tan(x)=c$.

To solve $\tan(x)=c$, we first determine one solution $x=\tan^{-1}(c)$. Then the general solution is given by

$$\boxed{x=\tan^{-1}(c)+n\cdot \pi} \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

Example $\PageIndex{2}$

Solve for $x$:

1. $\tan(x)=1$
2. $\tan(x)=-1$
3. $\tan(x)=5.1$
4. $\tan(x)=-3.7$

Solution

1. First, we find $\tan^{-1}(1)=\dfrac{\pi}{4}$. The general solution is thus:

$$x=\dfrac{\pi}{4}+n\cdot \pi \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

2. First, we need to find $\tan^{-1}(-1)$. Recall from equation [EQU:tan-1(-x)] that $\tan^{-1}(-c)=-\tan^{-1}(c)$, and recall further that $\tan^{-1}(1)=\dfrac{\pi}{4}$. With this we have

$$\tan^{-1}(-1)=-\tan^{-1}(1)=-\dfrac{\pi}{4} \nonumber$$

The general solution of $\tan(x)=-1$ is therefore,

$$x=-\dfrac{\pi}{4}+n\cdot \pi,\quad \text{ where } n=0,\pm 1, \pm 2, \dots \nonumber$$

For parts (c) and (d), we do not have an exact solution, so that the solution can only be approximated with the calculator.

3. $x=\tan ^{-1}(5.1)+n \pi \quad \approx 1.377+n \pi, \quad \text { where } n=0, \pm 1, \pm 2, \ldots$
4. $x=\tan ^{-1}(-3.7)+n \pi \quad \approx-1.307+n \pi, \quad \text { where } n=0, \pm 1, \pm 2, \ldots$

The equation $\cos(x)=c$

We start again with an example.

Example $\PageIndex{3}$

Solve for $x$: $\cos(x)=\dfrac{1}{2}$

Solution

We have the obvious solution to the equation $x=\cos^{-1}\left(\dfrac 1 2\right)=\dfrac{\pi}{3}$. However, since $\cos(-x)=\cos(x)$, there is another solution given by taking $x=-\dfrac{\pi}{3}$:

$$\cos\Big(-\dfrac{\pi}{3}\Big)=\cos\Big(\dfrac{\pi}{3}\Big)=\dfrac{1}{2} \nonumber$$

Moreover, the $y=\cos(x)$ function is periodic with period $2\pi$, that is, we have $\cos(x+2\pi)=\cos(x)$. Thus, all of the following numbers are solutions of the equation $\cos(x)=\dfrac 1 2$:

$$\begin{array}{rrrrrrr} & \ldots, & \dfrac{\pi}{3}-4 \pi, & \dfrac{\pi}{3}-2 \pi, & \dfrac{\pi}{3}, & \dfrac{\pi}{3}+2 \pi, & \dfrac{\pi}{3}+4 \pi, & \ldots, \\ \text { and: } & \ldots, & -\dfrac{\pi}{3}-4 \pi, & -\dfrac{\pi}{3}-2 \pi, & -\dfrac{\pi}{3}, & -\dfrac{\pi}{3}+2 \pi, & -\dfrac{\pi}{3}+4 \pi, & \ldots \end{array} \nonumber$$

From the graph we see that there are only two solutions of $\cos(x)=\dfrac 1 2$ within one period. Thus, the above list constitutes all solutions of the equation. With this observation, we may write the general solution as:

$$x=\dfrac{\pi}3+2n\cdot \pi, \text{ or } x=-\dfrac{\pi}{3}+2n\cdot \pi, \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots$$

In short, we write this as: $x=\pm\dfrac{\pi}3+2n\cdot \pi$ with $n=0, \pm 1, \pm 2, \pm 3, \dots$.

We generalize this example as follows.

Observation: Solve $cos(x)=c$

To solve $\cos(x)=c$, we first determine one solution $x=\cos^{-1}(c)$. Then the general solution is given by

$$x=\cos^{-1}(c)+2n\cdot \pi, \text{ or } x=-\cos^{-1}(c)+2n\cdot \pi, \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

In short, we write

$$\boxed{x=\pm\cos^{-1}(c)+2n\cdot \pi } \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

Example $\PageIndex{4}$

Solve for $x$.

1. $\cos(x)=-\dfrac{\sqrt{2}}{2}$
2. $\cos(x)=0.6$
3. $\cos(x)=-3$
4. $\cos(x)=-1$

Solution

1. First, we need to find $\cos^{-1}\left(-\dfrac{\sqrt{2}}{2}\right)$. From equation [EQU:cos-1(-x)] we know that $\cos^{-1}(-c)=\pi-\cos^{-1}(c)$, so that:

$$\cos^{-1}\bigg(-\dfrac{\sqrt{2}}{2}\bigg)=\quad \pi-\cos^{-1}\bigg(\dfrac{\sqrt{2}}{2}\bigg)= \quad \pi- \dfrac{\pi}{4}=\quad \dfrac{4\pi-\pi}{4}=\quad \dfrac{3\pi}{4}$$

The solution is therefore,

$$x=\pm\dfrac{3\pi}{4}+2n\pi, \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

2. We calculate $\cos^{-1}(0.6)\approx 0.927$ with the calculator. The general solution is therefore,

$$x=\pm\cos^{-1}(0.6)+2n\pi\quad\approx \pm 0.927+2n\pi, \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

3. Since the cosine is always $-1\leq\cos(x)\leq 1$, the cosine can never be $-3$. Therefore, there is no solution to the equation $\cos(x)=-3$. This can also be seen from the graph which does not intersect with the horizontal line $y=-3$.

4. A special solution of $\cos(x)=-1$ is

$$\cos^{-1}(-1)=\pi-\cos^{-1}(1)=\pi-0=\pi \nonumber$$

so that the general solution is

$$x=\pm\pi+2n\pi, \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

However, since $-\pi+2\pi=+\pi$, the solutions $\pi+2n\pi$ and $-\pi+2n\pi$ (for $n=0,\pm1,\pm 2, \dots$) can be identified with each other, and there is only one solution in each period. Thus, the general solution can be written as

$$x=\pi+2n\pi, \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

The graphs of $y=\cos(x)$ and $y=-1$ confirm these considerations.

The equation $\sin(x)=c$

Example $\PageIndex{5}$

Solve for $x$: $\sin(x)=\dfrac{\sqrt{2}}{2}$

Solution

First, we can find one obvious solution $x=\sin^{-1}\left(\dfrac{\sqrt{2}}{2}\right)=\dfrac{\pi}{4}$. Furthermore, from the top right equation in [EQU:basic-trig-eqns-wrt-pi], we have that $\sin(\pi-x)=\sin(x)$, so that another solution is given by $\pi-\dfrac{\pi}{4}$:

$$\sin\left(\pi-\dfrac{\pi}{4}\right)=\sin\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2} \nonumber$$

(This can also be seen by going back to the unit circle definition.) These are all solutions within one period, as can be checked from the graph above. The function $y=\sin(x)$ is periodic with period $2\pi$, so that adding $2n\cdot \pi$ for any $n=0,\pm 1, \pm 2, \dots$ gives all solutions of $\sin(x)=\dfrac{\sqrt{2}}{2}$. This means that the general solution is:

$$x=\dfrac{\pi}{4}+2n\cdot \pi, \text{ or }x=\left(\pi-\dfrac{\pi}{4}\right)+2n\cdot \pi, \quad \text{ for } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

We rewrite these solutions to obtain one single formula for the solutions. Note, that $\pi-\dfrac{\pi}{4}+2n\cdot \pi=-\dfrac{\pi}{4}+(2n+1)\cdot \pi$. Therefore, all solutions are of the form

$$x= \pm\dfrac{\pi}{4}+k\cdot \pi \nonumber$$

where, for even $k=2n$, the sign in front of $\dfrac{\pi}{4}$ is “$+$,” and, for odd $k=2n+1$, the sign in front of $\dfrac{\pi}{4}$ is “$-$.” This can be summarized as:

$$x=(-1)^k\cdot \dfrac{\pi}{4}+k\cdot \pi \nonumber$$

Renaming the indexing variable $k$ to the usual $n$, we obtain the final version for the solutions of $\sin(x)=\dfrac{\sqrt{2}}{2}$.

$$x=(-1)^n\cdot\dfrac{\pi}{4}+n\cdot \pi, \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

We have the following general statement.

Observation: Solve $sin(x)=c$

To solve $\sin(x)=c$, we first determine one solution $x=\sin^{-1}(c)$. Then the general solution is given by

$$\boxed{x=(-1)^n\cdot \sin^{-1}(c)+n\cdot \pi} \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots$$

Example $\PageIndex{6}$

Solve for $x$.

1. $\sin(x)=\dfrac{1}{2}$
2. $\sin(x)=-\dfrac{1}{2}$
3. $\sin(x)=-\dfrac{5}{7}$
4. $\sin(x)=1$

Solution

1. First, we calculate $\sin^{-1}\left(\dfrac 1 2\right)=\dfrac{\pi}{6}$. The general solution is therefore,

$$x=(-1)^n\cdot\dfrac{\pi}{6}+n\cdot \pi, \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

2. First, we need to find $\sin^{-1}\left(-\dfrac 1 2\right)$. From equation [EQU:sin-1(-x)] we know that $\sin^{-1}(-c)=-\sin^{-1}(c)$, so that $\sin^{-1}\left(-\dfrac 1 2\right)=-\sin^{-1}\left(\dfrac 1 2\right)=-\dfrac{\pi}{6}$. We thus state the general solution as

$$x=(-1)^{n} \cdot\left(-\dfrac{\pi}{6}\right)+n \cdot \pi=-(-1)^{n} \cdot \dfrac{\pi}{6}+n \cdot \pi=(-1)^{n+1} \cdot \dfrac{\pi}{6}+n \cdot \pi , \text{ where } n=0, \pm 1, \pm 2, \pm 3, \ldots \nonumber$$

3. We do not have an exact value of $\sin^{-1}\left(-\dfrac 5 7\right)$, so that we either need to leave it as this, or approximate this with the calculator $\sin^{-1}\left(-\dfrac 5 7\right)\approx -0.796$. We get the solution:

$$x \approx(-1)^{n} \cdot(-0.796)+n \cdot \pi=(-1)^{n+1} \cdot 0.796+n \cdot \pi , \text{ where } n=0, \pm 1, \pm 2, \pm 3, \ldots \nonumber$$

4. We calculate $\sin^{-1}(1)=\dfrac{\pi}{2}$. The general solution $x$ can be written as

$$\label{EQU:a-trig-example} x=(-1)^n\cdot\dfrac{\pi}{2}+n\cdot \pi, \quad \text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots$$

Now, if we look at the graph, we see that each period only has one solution.

Algebraically, we can see this as follows. For an even number $n$, the solution $x=(-1)^n\cdot\dfrac{\pi}{2}+n\cdot \pi$ coincides with the solution coming from the index $n+1$, that is:

$$\begin{aligned} (-1)^n\cdot\dfrac{\pi}{2}+n\cdot \pi &= +\dfrac{\pi}{2}+n\cdot \pi, \text{ and } \\ (-1)^{n+1}\cdot\dfrac{\pi}{2}+(n+1)\cdot \pi &= -\dfrac{\pi}{2}+(n+1)\cdot \pi = -\dfrac{\pi}{2}+n\cdot \pi +\pi\\ &= +\dfrac{\pi}{2}+n\cdot \pi\end{aligned} \nonumber$$

Therefore, we can write the solution more efficiently by removing the odd solutions (since they coincide with the even solutions), and state this as

$$x=(-1)^n\cdot\dfrac{\pi}{2}+n\cdot \pi, \quad \text{ where } n=0, \pm 2, \pm 4, \dots \nonumber$$

Since $(-1)^n=+1$ for even $n$, we can just write this as

$$\label{EQU:a-trig-example-simplified} x=\dfrac{\pi}{2}+2n\cdot \pi, \quad \text{ where } n=0, \pm 1, \pm 2, \pm 3, \dots$$

Writing the solutions as $x=\dfrac{\pi}{2}+2n\cdot \pi$ as in $\ref{EQU:a-trig-example-simplified}$ instead of the original $x=(-1)^n\cdot\dfrac{\pi}{2}+n\cdot \pi$ from $\ref{EQU:a-trig-example}$ for $n=0,\pm1, \pm2, \dots$ certainly does not change the overall solution set. However, writing the solutions as in $\ref{EQU:a-trig-example-simplified}$ is more efficient, since it does not repeat any of the solutions, and is therefore a simplified and preferred way of presenting the solutions.

Summary

We summarize the different formulas we used to solve the basic trigonometric equations in the following table.

Solve: $\sin (x)=c$	Solve: $\cos (x)=c$	Solve: $\tan (x)=c$
First, find one solution, that is: $\sin ^{-1}(c)$ Use: $\sin ^{-1}(-c)=-\sin ^{-1}(c)$	First, find one solution, that is: $\cos ^{-1}(c)$ Use: $\cos ^{-1}(-c)= \pi -\cos ^{-1}(c)$	First, find one solution, that is: $\tan ^{-1}(c)$ Use: $\tan ^{-1}(-c)=-\tan ^{-1}(c)$
The general solution is: $\boxed {x=(-1)^{n} \sin ^{-1}(c)+n \pi}$ where $n=0, \pm 1, \pm 2, \ldots$	The general solution is: $\boxed {x=\pm \cos ^{-1}(c)+2 n \pi}$ where $n=0, \pm 1, \pm 2, \ldots$	The general solution is: $\boxed {x=\tan ^{-1}(c)+n \pi}$ where $n=0, \pm 1, \pm 2, \ldots$

Example $\PageIndex{7}$

Find the general solution of the equation, and state at least $5$ distinct solutions.

1. $\sin(x)=-\dfrac{1}{2}$
2. $\cos(x)=-\dfrac{\sqrt{3}}{2}$

Solution

1. We already calculated the general solution in example $\PageIndex{6}$ (b). It is

$$x=(-1)^{n+1}\cdot\dfrac{\pi}{6}+n\cdot \pi, \quad \text{where } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

We simplify the solutions for $n=0,1,-1,2,-2$.

$$\begin{aligned} n=0: x&= (-1)^{0+1}\cdot\dfrac{\pi}{6}+0\cdot \pi =-\dfrac{\pi}{6} \\ n=1: x&= (-1)^{1+1}\cdot\dfrac{\pi}{6}+1\cdot \pi =\dfrac{\pi}{6}+\pi =\dfrac{\pi+6\pi}{6}=\dfrac{7\pi}{6} \\ n=-1: x&= (-1)^{-1+1}\cdot\dfrac{\pi}{6}+(-1)\cdot \pi =\dfrac{\pi}{6}-\pi =\dfrac{\pi-6\pi}{6}=\dfrac{-5\pi}{6} \\ n=2: x&= (-1)^{2+1}\cdot\dfrac{\pi}{6}+2\cdot \pi =-\dfrac{\pi}{6}+2\pi =\dfrac{-\pi+12\pi}{6}=\dfrac{11\pi}{6} \\ n=-2: x&= (-1)^{-2+1}\cdot\dfrac{\pi}{6}+(-2)\cdot \pi =-\dfrac{\pi}{6}-2\pi =\dfrac{-\pi-12\pi}{6}=\dfrac{-13\pi}{6}\end{aligned} \nonumber$$

2. It is $\cos^{-1}\left(-\dfrac{\sqrt{3}}{2}\right)=\pi-\cos^{-1}\left(\dfrac{\sqrt{3}}{2}\right)=\pi-\dfrac{\pi}{6}=\dfrac{6\pi-\pi}{6}=\dfrac{5\pi}{6}$. The solutions of $\cos(x)=-\dfrac{\sqrt{3}}{2}$ are:

$$x=\pm\dfrac{5\pi}{6}+2n\pi, \quad\text{ where } n=0,\pm 1, \pm 2, \pm 3, \dots \nonumber$$

We write the $6$ solutions with $n=0,+1,-1$, and for each use the two distinct first terms $+\dfrac{5\pi}6$ and $-\dfrac{5\pi}6$.

$$\begin{aligned} n=0: x&= +\dfrac{5\pi}6+2\cdot 0 \cdot \pi=\dfrac{5\pi}6 \\ n=0: x&= -\dfrac{5\pi}6+2\cdot 0 \cdot \pi=-\dfrac{5\pi}6 \\ n=1: x&= +\dfrac{5\pi}6+2\cdot 1 \cdot \pi=\dfrac{5\pi}6+2\pi=\dfrac{5\pi+12\pi}{6}=\dfrac{17\pi}{6} \\ n=1: x&= -\dfrac{5\pi}6+2\cdot 1 \cdot \pi=-\dfrac{5\pi}6+2\pi=\dfrac{-5\pi+12\pi}{6}=\dfrac{7\pi}{6} \\ n=-1: x&= +\dfrac{5\pi}6+2\cdot (-1) \cdot \pi=\dfrac{5\pi}6-2\pi=\dfrac{5\pi-12\pi}{6}=\dfrac{-7\pi}{6} \\ n=-1: x&= -\dfrac{5\pi}6+2\cdot (-1) \cdot \pi=-\dfrac{5\pi}6-2\pi=\dfrac{-5\pi-12\pi}{6}=\dfrac{-17\pi}{6}\end{aligned} \nonumber$$

Further solutions can be found by taking values $n=+2,-2,+3,-3,\dots$.