Solving Trigonometric Equations

Problem 1. Determine the value(s) of $x$ such that $\sin(x) = 1/2$ , where

$0 < x < \pi/2$ ,
$0 \leq x < 2\pi$ ,
$x$ is a real number.

We call this process solving a trigonometric equation.

(Click for Solution)

Solution. Using special angles, we recall that for $0 < x < \pi/2$ ,

$\sin(x) = 1/2 \quad \iff \quad x = \pi/6.$

Denote $\alpha := \pi/6$ . Using the extended definitions of $\sin(x)$ , for $0 \leq x < 2\pi$ , we have

$\begin{aligned} \sin(\alpha) &= 1/2,\\ \sin(\pi - \alpha) &= \sin(\alpha) = 1/2,\\ \sin(\pi + \alpha) &= -{ \sin(\alpha) } = -1/2,\\ \sin(2\pi - \alpha) &= -{ \sin(\alpha) } = -1/2. \end{aligned}$

Since only the first two equations work, we have $x = \pi/6$ or

$x = \pi - \pi/6 = 5\pi/6.$

Graphically, we have the following.

Finally, for general $x$ , we recall that $\sin(x)$ is $2\pi$ -periodic:

$\sin(x) = \sin(x+2\pi).$

By repeating this process, given any integer $k$ , we have

$\sin(x + 2k\pi) = \sin(x).$

In particular,

$\sin(\pi/6 + 2k\pi) = \sin(\pi/6) = 1/2.$

Furthermore, for any $x_0 \neq \pi/6$ and $x_0 \neq 5\pi/6$ ,

$\sin(x_0 + 2k\pi) = \sin(x_0) \neq 1/2.$

Therefore, we must have $x = \pi/6 + 2k\pi$ or $x = 5\pi/6 + 2k\pi$ , where $k$ is some integer.

Problem 2. Solve the equation $\cos(x) = 1/2$ , where

$0 < x < \pi/2$ ,
$0 \leq x < 2\pi$ .

(Click for Solution)

Solution. Using special angles, we recall that for $0 < x < \pi/2$ ,

$\cos(x) = 1/2 \quad \iff \quad x = \pi/3.$

Using the extended definitions of $\cos(x)$ , for $0 \leq x < 2\pi$ , we have

$\begin{aligned} \cos(\pi/3) &= 1/2,\\ \cos(\pi - \pi/3) &= -{ \cos(\pi/3) } = -1/2,\\ \cos(\pi + \pi/3) &= -{ \cos(\pi/3) } = -1/2,\\ \cos(2\pi - \pi/3) &= \cos(\pi/3) = 1/2. \end{aligned}$

Since only the first and last equations work, we have $x = \pi/3$ or

$x = 2\pi - \pi/3 = 5\pi/3.$

Graphically, we have the following.

Problem 3. Solve the equation $\tan(x) = 1$ , where $0 \leq x < 2\pi$ .

(Click for Solution)

Solution. Using special angles, we recall that for $0 < x < \pi/2$ ,

$\tan(x) = 1 \quad \iff \quad x = \pi/4.$

Using the extended definitions of $\tan(x)$ , for $0 \leq x < 2\pi$ , we have

$\begin{aligned} \tan(\pi/4) &= 1,\\ \tan(\pi - \pi/4) &= -{ \tan(\pi/4) } = -1,\\ \tan(\pi + \pi/4) &= \tan(\pi/4) = 1, \\ \tan(2\pi - \pi/4) &= -{ \tan(\pi/4) } = -1. \end{aligned}$

Since only the first and third equations work, we have $x = \pi/4$ or

$x = \pi + \pi/4 = 5\pi/4.$

Graphically, we have the following.

Problem 4. For each of the cases $r=-1, 0, 1$ , solve the equation $\sin(x) = r$ , where $0 \leq x \leq 2\pi$ .

(Click for Solution)

Solution. Firstly, we work with the case $r = 1$ . Using special angles, we recall that for $0 \leq x \leq \pi/2$ ,

$\sin(x) = 1 \quad \iff \quad x = \pi/2.$

Using the extended definitions of $\sin(x)$ , for $0 \leq x \leq 2\pi$ , we have

$\begin{aligned} \sin(\pi/2) &= 1,\\ \sin(\pi - \pi/2) &= \sin(\pi/2) = 1,\\ \sin(\pi + \pi/2) &= -{ \sin(\pi/2) } = -1, \\ \sin(2\pi - \pi/4) &= -{ \sin(\pi/2) } = -1. \end{aligned}$

Since only the first and third equations work, we have $x = \pi/2$ or

$x = \pi - \pi/2 = \pi/2.$

In fact, we get one and only one solution: $x = \pi/2$ .

Secondly, we work with the case $r = 0$ . Using special angles, we recall that for $0 \leq x \leq \pi/2$ ,

$\sin(x) = 0 \quad \iff \quad x = 0.$

Using the extended definitions of $\sin(x)$ , for $0 \leq x \leq 2\pi$ , we have

$\begin{aligned} \sin(0) &= 0,\\ \sin(\pi - 0) &= \sin(0) = 0,\\ \sin(\pi + 0) &= -{ \sin(0) } = 0, \\ \sin(2\pi - 0) &= -{ \sin(0) } = 0. \end{aligned}$

Now, all equations work, so that we have exactly three solutions: $x = 0, \pi, 2\pi$ .

Finally, we work with the case $r = -1$ . Using the same equations as the case for $r = 1$ , the solutions must be

$x = \pi + \pi/2 = 3\pi/2$ , or
$\quad x = 2\pi - \pi/2 = 3\pi/2$ .

In fact, we get one and only one solution: $x = 3\pi/2$ .

Notice that the $x$ -values of interest take the form $k\pi/2$ , where $0 \leq k \leq 4$ is an integer.

Graphically, we have the following.

Problem 5. For each of the cases $r=-1, 0, 1$ , solve the equation $\cos(x) = r$ , where $0 \leq x \leq 2\pi$ .

(Click for Solution)

Solution. We can either adopt the strategy in Problem 4, or simply focus on the special values $k\pi/2$ , where $0 \leq k \leq 4$ is an integer. Observe that

$\begin{aligned} \cos(0) &= \cos(2\pi) = 1, \\ \cos(\pi/2) &= \cos(3\pi/2) = 0, \\ \cos(\pi) &= -1. \end{aligned}$

Therefore, for $0 \leq x \leq 2\pi$ , we must have:

$\cos(x) = 1 \iff x = 0, 2\pi$ ,
$\cos(x) = 0 \iff x = \pi/2, 3\pi/2$ ,
$\cos(x) = -1 \iff x = \pi$ .

Graphically, we have the following.

Problem 6. Fix $-1 \leq r \leq 1$ . Define $0 \leq \alpha \leq \pi/2$ such that

$\sin(\alpha) = |r|.$

Solve the equation $\sin(x) = r$ in terms of $\alpha$ , where $0 \leq x \leq 2\pi$ .

(Click for Solution)

Solution. Using the extended definitions of $\sin(x)$ , for $0 \leq x \leq 2\pi$ , we have

$\begin{aligned} \sin(\alpha) &= |r|,\\ \sin(\pi - \alpha) &= \sin(\alpha) = |r|,\\ \sin(\pi + \alpha) &= -{ \sin(\alpha) } = -|r|,\\ \sin(2\pi - \alpha) &= -{ \sin(\alpha) } = -|r|. \end{aligned}$

We consider the three cases:

If $r > 0$ , then only the first two equations work: $x = \alpha, \pi -\alpha$ .
If $r = 0$ , then $\alpha = 0$ : $x = 0, \pi, 2\pi$ by Problem 4.
If $r < 0$ , then only the last two equations work: $x = \pi + \alpha, 2\pi -\alpha$ .

Graphically, we have the following.

Problem 7. Fix $-1 \leq r \leq 1$ . Define $0 \leq \alpha \leq \pi/2$ such that

$\cos(\alpha) = |r|.$

Solve the equation $\cos(x) = r$ in terms of $\alpha$ , where $0 \leq x \leq 2\pi$ .

(Click for Solution)

Solution. Using the extended definitions of $\cos(x)$ , for $0 \leq x \leq 2\pi$ , we have

$\begin{aligned} \cos(\alpha) &= |r|,\\ \cos(\pi - \alpha) &= -{ \cos(\alpha) } = -|r|,\\ \cos(\pi + \alpha) &= -{ \cos(\alpha) } = -|r|,\\ \cos(2\pi - \alpha) &= { \cos(\alpha) } = |r|. \end{aligned}$

We consider the three cases:

If $r > 0$ , then only the first and last equations work: $x = \alpha, 2\pi -\alpha$ .
If $r = 0$ , then $\alpha = 0$ : $x = \pi/2, 3\pi/2$ by Problem 5.
If $r < 0$ , then only the middle two equations work: $x = \pi - \alpha, \pi +\alpha$ .

Graphically, we have the following.

Problem 8. Fix any real number $r$ . Define $0 \leq \alpha \leq \pi/2$ such that

$\tan(\alpha) = |r|.$

Solve the equation $\tan(x) = r$ in terms of $\alpha$ , where $0 \leq x \leq 2\pi$ .

(Click for Solution)

Solution. Using the extended definitions of $\tan(x)$ , for $0 \leq x \leq 2\pi$ , we have

$\begin{aligned} \tan(\alpha) &= |r|,\\ \tan(\pi - \alpha) &= -{ \tan(\alpha) } = -|r|,\\ \tan(\pi + \alpha) &= { \tan(\alpha) } = |r|,\\ \tan(2\pi - \alpha) &= { \tan(\alpha) } = -|r|. \end{aligned}$

We consider the three cases:

If $r > 0$ , then only the first and last equations work: $x = \alpha, \pi + \alpha$ .
If $r < 0$ , then only the middle two equations work: $x = \pi - \alpha, 2\pi -\alpha$ .
If $r = 0$ , then $\alpha = 0$ : $x = 0, \pi, 2\pi$ .

Graphically, we have the following.

Remark 1. Problems 6–8 detail the ASTC strategy used to solve trigonometric equations. Given a trigonometric equation $f(x) = r$ , the solutions for $0 \leq x \leq 2\pi$ depend on

the type of trigonometric function $f(x)$ , and
the sign of $r$ .

Example 1. Given the equation $\sin(x) = r$ , by calculating $0 \leq \alpha \leq \pi/2$ such that $\sin(\alpha) = |r|$ , the candidate solutions in the domain $0 \leq x < 2\pi$ for the equation are

$x = \alpha, \quad x = \pi-\alpha,\quad x = \pi+ \alpha,\quad x = 2\pi -\alpha.$

If $r \geq 0$ , we choose the first two candidates (i.e. ASTC), yielding

$x = \alpha, \quad x = \pi-\alpha.$

If $r < 0$ , we choose the latter two candidates (i.e. ASTC), yielding

$x = \pi + \alpha, \quad x = 2\pi-\alpha.$

Graphically, we have the following.

We then add and subtract enough angles by multiples of $2\pi$ until we recover all desired solutions, as per the solving strategy in Problems 1 and 3.

Example 2. Given the equation $\cos(x) = r$ , by calculating $0 \leq \alpha \leq \pi/2$ such that $\cos(\alpha) = |r|$ , the candidate solutions in the domain $0 \leq x < 2\pi$ for the equation are

$x = \alpha, \quad x = \pi-\alpha,\quad x = \pi+ \alpha,\quad x = 2\pi -\alpha.$

If $r \geq 0$ , we choose the first and last candidates (i.e. ASTC), yielding

$x = \alpha, \quad x = 2\pi-\alpha.$

If $r < 0$ , we choose the middle two candidates (i.e. ASTC), yielding

$x = \pi - \alpha, \quad x = \pi + \alpha.$

We then add and subtract enough angles by multiples of $2\pi$ until we recover all desired solutions, as per the solving strategy in Problems 2 and 5.

Graphically, we have the following.

Example 3. Given the equation $\tan(x) = r$ , by calculating $0 \leq \alpha \leq \pi/2$ such that $\tan(\alpha) = |r|$ , the candidate solutions in the domain $0 \leq x < 2\pi$ for the equation are

$x = \alpha, \quad x = \pi-\alpha,\quad x = \pi+ \alpha,\quad x = 2\pi -\alpha.$

If $r \geq 0$ , we choose the first and third candidates (i.e. ASTC), yielding

$x = \alpha, \quad x = \pi+\alpha.$

If $r < 0$ , we choose the second and fourth candidates (i.e. ASTC), yielding

$x = \pi - \alpha, \quad x = 2\pi - \alpha.$

Graphically, we have the following.

We then add and subtract enough angles by multiples of $2\pi$ until we recover all desired solutions, as per the solving strategy in Problems 2 and 5.

Problem 9. Solve the equation $\sin(2x) = 1/2$ for $0 \leq x \leq 2\pi$ .

(Click for Solution)

Solution. Write $u = 2x$ so that $0 \leq u \leq 4\pi$ . To solve the equation

$\sin(u) = 1/2,$

we appeal to Problem 1 (or ASTC techniques) to conclude that for $0 \leq u \leq 2\pi$ ,

$u = \pi/6,\quad 5\pi/6.$

To account for all solutions $0 \leq u \leq 4\pi$ , we add $2\pi$ to each solution:

$u = \pi/6,\quad 5\pi/6,\quad 13\pi/6,\quad 17\pi/6.$

Since $u = 2x$ , we divide by $2$ to obtain all solutions in $x$ :

$x = \pi/12,\quad 5\pi/12,\quad 13\pi/12,\quad 17\pi/12.$

Problem 10. Solve the equation $\cos(4x) - 3\sin(2x) + 1 = 0$ for $0 \leq x \leq 2\pi$ .

Hint. Use the double-angle formula $\cos(2A) = 1 - 2 \sin^2(A)$ .

(Click for Solution)

Solution. Using the double-angle formula,

$\cos(4x) = \cos(2 \cdot 2x) = 1 - 2 \sin^2(2x).$

Hence, we can simplify the given equation in terms of $\sin(2x)$ :

$\begin{aligned} \cos(4x) - 3\sin(2x) + 1 &= 0 \\ (1 - 2 \sin^2(2x)) - 3\sin(2x) + 1 &= 0 \\ - 2 \sin^2(2x) - 3\sin(2x) + 2 &= 0 \\ 2\sin^2(2x) + 3\sin(2x) - 2 &= 0 \\ (\sin(2x) + 2)(2\sin(2x) - 1) &= 0. \end{aligned}$