Trigonometric Equations – Explanation and Examples

Trigonometric equations are equations that involve trigonometric functions and unknown angles.

These equations often require the use of trigonometric identities to solve for the unknown variable. Since the trigonometric functions are periodic, there is a solution between $0$ and $2\pi$ if a solution exists, but infinitely many more solutions will exist if so. It is also possible that there is no solution to the equation.

Since trigonometric equations involve solving for unknown variables, they are useful in all concrete applications of trigonometry, including science and engineering.

Make sure to review how to solve linear and quadratic equations, inverse trig functions, and trig identities before reading this section.

What Is a Trigonometric Equation?

A trigonometric equation is an equation involving trigonometric functions and unknown angles. These unknown angles will be the variables in the equations, and since trigonometric functions are periodic, they can have infinitely many solutions only if a real-valued solution exists.

Usually, the solution to an equation is assumed to be a solution on the interval $0$ to $2\pi$ radians unless otherwise specified. This is called the principal solution. Sometimes, however, a general solution is requested. This is a solution that includes all of the possible angles that could work in the equation. Such a solution will include a variable “n” where “n” is an integer. 

A simple example of a trigonometric equation is $sinx=0$. Since sine is equal to $0$ at $0$ radians and $\pi$ radians, a principal solution is $0$ or $\pi$. The general solution is $0+n\pi$, where $n$ is an integer. 

Most of the time, however, trigonometric equations will require more work than simply using the inverse trig functions. This is because they are usually a little more complex. Typically, using trigonometric identities to rewrite the equation using only one type of trigonometric function is the best way to solve trigonometric equations.

How To Solve a Trigonometric Equation

To solve a trigonometric equation, the trig functions should be treated like variables to an extent, even though they are functions of the variable. Solving trigonometric equations works in much the same way as solving polynomial equations when there is only one trig function involved.

If the highest power of a trig equation is $1$, get all of the trig functions on one side of the equation and everything else on the other side of the equation. Then, use the inverse trig functions on both sides of the equation to cancel the trig function. This will create a general polynomial equation, which can be solved using reverse order of operations.

If the highest power of the trig functions is greater than $1$, it can be solved in the same way as polynomials with powers greater than $1$. These are easiest to solve when they can be factored as a quadratic. Factoring by grouping also works when the power is greater than $2$.

If there is more than one trig function, however, it is necessary to first rewrite the equation so there is only one function. This is done using trigonometric identities. After that, these steps can be used to simplify and solve the equation.

Trigonometric Equations With More Than One Trig Function

Sometimes trigonometric equations will involve more than one trigonometric function. In this case, it does not make sense to apply an inverse trig function to both sides of the equation since that will not cancel all of the trig functions. In this case, use trigonometric identities to make all of the trig functions the same. 

If a trig function is added or subtracted (such as $sinx+siny$), use the addition/subtraction identities to compose this into one function.

The ultimate goal of solving trig functions is to get a single trig function in the first degree on one side of the equation and numbers on the other side. Doing this means that the left side can be canceled through an inverse trig function, which makes solving for the variable possible if a solution exists.

Basic Trigonometric Equations

Basic trigonometric equations have the form $funx=a$, where $fun$ stands for any of the six trigonometric functions, sine, cosine, tangent, cosecant, secant, and cotangent, and $a$ is a number. 

These are basic trigonometric equations because they do not require any rearranging to solve. To cancel the trigonometric function, apply the inverse trigonometric functions to both sides. Therefore, the solution to this equation is $x=arcfun(a)$, where $arcfun$ is the corresponding inverse trig function.

When solving an equation of the form $fun(f(x))=a$, begin as before by applying the inverse trigonometric function to both sides. This yields $f(x)=arcfun(a)$. Then, solve for $x$ using algebraic methods.

If the trig function is sine, cosine, secant, or cosecant, the general solution is $arcfun(a)+2n\pi$ for any integer $n$. If the trig function is tangent or cotangent, the general solution is $arcfun(a)+n\pi$ for any integer $n$.

Knowing how to solve these basic equations is important because the ultimate goal with general trigonometric equation solving is getting the equation into the form $fun(x)=a$ or $fun(f(x))=a$.

Trigonometric Equations With No Solution

Not all trigonometric equations have a solution. Some trigonometric equations do not have any solutions. This is because the inverse trig function is not always defined for all values.

For example, the sine and cosine of an angle must be between $-1$ and $1$. The cosecant and secant cannot be in the interval $(-1, 1)$.   

Therefore, equations like $cosx=5$ and $cscx=0$ have no solution in the real numbers. There are solutions involving imaginary (complex) numbers, but they are outside the scope of this article.

Take note that although tangent and cotangent are defined for all real values, it is still possible to end up with an equation where $tanx=c$ for a complex, not-real number $c$. Such a function would also have no real-valued solution.

Examples

This section goes over common problems involving trigonometric equations and their step-by-step solutions.

Example 1

Find the principal and general solutions for the basic trigonometric equation $cosx=-1$.

Solution

This equation is already in the form $funx=a$. Therefore, to solve this, apply the inverse trigonometric function — in this case, the arccosine function — to both sides of the equation.

$arccos(cosx)=arccos(-1)$.

Simplifying, this is:

$x=\pi$ radians.

This is the principal solution, or the solution on the interval $(0, 2\pi)$. Since cosine is $2\pi$ periodic, however, the general solution is $\pi+2n\pi$ radians for any integer $n$. 

Example 2

Find the principal and general solutions to the trigonometric equation $sin^2x=1$.

Solution

Take note that this trigonometric equation looks a lot like the polynomial equation $x^2=1$. To solve this polynomial equation, take the square root of both sides. Solving the trigonometric equation $sin^2x=1$ begins in the same way.

$\sqrt{sin^2x} = \pm \sqrt{1}$.

Remember that there is a positive and a negative square root because $1\time1=1$ and $-1 \times -1=1$.

Therefore, simplifying this is:

$sinx=1$ or $sinx=-1$.

To solve this, take the arcsine of both sides of the two equations. This yields:

$x=\frac{\pi}{2}$ radians or $x=\frac{3\pi}{2}$ radians.

This is the principal solution. Since sine is $2\pi$ periodic, the general solution is $\frac{\pi}{2}+2n\pi$ radians or $\frac{3\pi}{2}+2n\pi$ radians for any integer $n$. This can be simplified as $\frac{\pi}{2}+n\pi$ radians for any integer $n$.

Example 3

Solve the trigonometric equation $tan(x^2+2x+1)+\sqrt{3} = 0$.

Solution

Unlike the previous examples, this trigonometric equation is not already in the form $fun(f(x))=a$ because the left side of the equation has both a trig function and a numeric term. To fix this, subtract $\sqrt{3}$ from both sides of the equation.

The new equation is:

$tan(x^2+2x+1) = -\sqrt{3}$.

Now, the equation is in the right form to take the inverse trig function of both sides. 

$arctan(tan(x^2+2x+1)) = arctan(-\sqrt{3})$.

This simplifies to:

$x^2+2x+1 = \frac{2\pi}{3}$.

But this doesn’t solve for the variable. Now, treat the equation like a normal quadratic. Move everything to the left side of the equation.

$x^2+2x+1-\frac{2\pi}{3}=0$.

Since the third term is not simple, use the quadratic formula to solve.

$x = \frac{-2 \pm \sqrt{4-4(1-\frac{2\pi}{3})}}{2}$ radians.

This simplifies to $x=1-\sqrt{\frac{2\pi}{3}}$ radians or $x = \sqrt{\frac{2\pi}{3}}-1$ radians, or approximately $-2.45$ radians or $0.45$ radians.

Since tangent is $\pi$ periodic, the general solutions are $x=1-\sqrt{\frac{2\pi}{3}}+n\pi$ radians or $x = \sqrt{\frac{2\pi}{3}}-1+n\pi$ radians for any integer $n$.

Example 4

Find the solution to the trigonometric equation $sin(x-1)+sin(x+1)+1=0$.

Solution

Begin by setting up the equation so the left side is only trig functions and the right side is only numbers. Doing this just requires subtracting $1$ from both sides.

$sin(x-1)+sin(x+1) = -1$.

This equation, however, is not of the form $funx=a$. There are two different sine functions. Use the sum formula to help.

$sin(x-1)+sin(x+1) = 2sin(\frac{x+1+x-1}{2})cos(\frac{x+1-(x-1)}{2})$.

This simplifies to:

$2sin(x)cos(1)$.

In this case, $cos(1)$ is just a number. It can be treated like a constant. Therefore, the trigonometric equation can be rewritten as:

$2sin(x)cos(1) = -1$ 

Or

$sin(x) = \frac{-1}{2cos(1)}$.

Take the arcsine of both sides to get the exact answer:

$x = arcsin(\frac{-1}{2cos(1)})$.

The decimal approximation of this is $-1.18$ radians. Therefore, the general solution is $-1.18+2n\pi$ for any integer $n$.

Example 5

Find the solution to the trigonometric equation $sin^2(2x)+2cos^2(2x) = -\frac{\sqrt{2}}{2}$.

Solution

Since this problem has two different types of trigonometric functions, it is necessary to use trigonometric identities so there is only one function. Then, it will be possible to solve for $x$.

In this case, there is a sine squared function and a cosine squared function, so it makes sense to consider the Pythagorean identities.

Specifically, $sin^2x+cos^2x=1$. This is true for any angle. Even though the inside of the trig functions in this case is $2x$, $2x$ is still just an angle. Therefore, $sin^2(2x)+cos^2(2x) = 1$ too.

Expanding the given equation yields $sin^2(2x)+cos^2(2x)+cos^2(2x) = -\frac{\sqrt{2}}{2}$.

The Pythagorean identity shows that this equation is equivalent to:

$1+cos^2(2x) = -\frac{\sqrt{2}}{2}$.

Subtracting $1$ from both sides makes this:

$cos^2(2x) = \frac{2-\sqrt{2}}{2}$.

Then, take the square root of both sides to get:

$cos(2x) = \sqrt{\frac{2-\sqrt{2}}{2}}$.

Now the equation is in the form $fun(f(x))=a$. Therefore, take the arccosine of both sides. This is:

$2x = arccos(\sqrt{\frac{2-\sqrt{2}}{2}}$.

Finally, divide both sides by $2$ to solve for $x$.

$x = \frac{arccos(\sqrt{\frac{2-\sqrt{2}}{2}}}{2}$ radians.

Although the right side of this equation looks complicated, it is just a constant. Plugging the values into a calculator reveals that it has the decimal approximation of 0.50 radians.

This is the principal solution. Since cosine is $2\pi$ periodic, the general solution is $\frac{arccos(\sqrt{\frac{2-\sqrt{2}}{2}}}{2} + 2n\pi$ radians for any integer $n$.

Example 6

Find the solution to the equation $tan(x)+cot(x) = \frac{1}{2}$.

Solution

Once again, this equation contains two different types of trigonometric functions. Using trigonometric identities to make it so that there is only one type of function will make the equation simpler.

In this case, begin with the quotient identities $tanx=\frac{sinx}{cosx}$ and $cotx=\frac{cosx}{sinx}$. This yields:

$\frac{sinx}{cosx}+\frac{cosx}{sinx} = \frac{1}{2}$.

Now multiply the first fraction by $\frac{sinx}{sinx}$ and the second by $\frac{cosx}{cosx}$. The two fractions have like denominators, and adding them together gives us:

$\frac{sin^2x+cos^2x}{cosxsinx} = \frac{1}{2}$.

The Pythagorean identity shows that the numerator is equal to $1$. Therefore, the whole equation simplifies to setting the denominators equal.

$cosxsinx=2$.

This looks similar to the identity $2cosxsinx=sin(2x)$. Therefore, multiply both sides of the equation by $2$.

$2cosxsinx=4$

$sin(2x)=4$.

But the sine of any angle must be between $-1$ and $1$. Therefore, there is no angle $2x$ such that $sin(2x)=4$. Thus, there is no solution to this equation.