# How to Solve Trigonometric Equations

Copyright © 2002–2024 by Stan Brown, BrownMath.com

Copyright © 2002–2024 by Stan Brown, BrownMath.com

**Summary:**
A **trigonometric equation** is one that involves one or
more of the six functions sine, cosine, tangent, cotangent, secant,
and cosecant. Some trigonometric equations, like
*x* = cos *x*, can be solved only numerically,
through successive approximations. But a great
many can be solved analytically — in “closed form”, an
exact solution in symbols — and this page shows
you how to do it in five steps.

If a trig equation can be solved analytically, these steps will do it:

- Put the equation in terms of one function of one angle.
- Write the equation as one trig function of an angle equals a constant.
- Write down the possible value(s) for the angle.
- If necessary, solve for the variable.
- Apply any restrictions on the solution.

On this page I’ll walk you through solving several trig equations using these steps, showing you every detail. Once you know and understand the steps, you’ll be able to work some more examples more quickly.

Trig equations, like any equations, are really about numbers, not angles. You are looking for all possible numbers that could be substituted for the variable in the equation to make it true. But it simplifies things to think about the angles first and worry about the variables later.

**Example A**:

cos(4*A*) − sin(2*A*) = 0

Here the “angles”, the arguments to the trig functions, are 4*A* and 2*A*.
True, you want to solve for *A* ultimately. But if you can solve for the
angle 4*A* or 2*A*, it is then quite easy to solve for
the variable.

As you see, that equation involves two functions (sine and cosine)
of two angles (4*A* and 2*A*). You need to get it in terms of one function
of one angle. Note well: *a function of one angle,* not
necessarily a function of just the variable *A*.

This is where it is essential to have
a nodding
acquaintance with all the trig identities. If you do, you’ll remember that
cos(2*u*) can be expressed in terms of sin(*u*).
Specifically,
cos(2*u*) = 1 − 2sin²(*u*).

How does that help? Well, 4*A* is 2×2*A*, isn’t it?

cos(2*u*) = 1 − 2sin²(*u*)

cos(2×2*A*) = 1 − 2sin²(2*A*)

cos(4*A*) = 1 − 2sin²(2*A*)

That transforms the original equation to

1 − 2sin²(2*A*) − sin(2*A*) = 0

which can be rewritten in standard form as

2sin²(2*A*) + sin(2*A*) − 1 = 0

Now you have the equation in terms of only one function (sine) and only
one angle (2*A*).

Now that the equation involves only a single function of a single angle, your next task is to solve for that function value.

**Example A** continues. Recapping what was done so far,

cos(4*A*) − sin(2*A*) = 0 ⇒

2sin²(2*A*) + sin(2*A*) − 1 = 0

You want to solve for sin(2*A*). You should recognize that the equation
is really a quadratic,

2*y*² + *y* − 1 = 0, where *y* = sin(2*A*)

It can be factored in a straightforward way,
as (*y*+1)(2*y*−1), which
gives

(sin(2*A*) + 1) (2sin(2*A*) − 1) = 0

From algebra you know that if a product is 0 then you solve by setting each factor to 0:

sin(2*A*) + 1 = 0 or 2sin(2*A*) − 1 = 0

sin(2*A*) = −1 or sin(2*A*) = 1/2

**Example B**:

This one is a bit simpler. Solve:

3tan²(*B*/2) − 1 = 0

To solve it, add 1 to both sides and divide by 3:

tan²(*B*/2) = 1/3

and then take square root of both sides:

tan(*B*/2) = ±√1/3 = ±√3/3

It’s important to remember to use the plus-or-minus sign ± when taking the square root of both sides; otherwise you could overlook some solutions.

After solving for a function value, now you solve for the angle. If it’s a multiple of π/6 (30°) or π/4 (45°), you can easily solve it exactly. If it’s a half-multiple of those angles, you may be able to use the half-angle formulas to write an exact solution. Otherwise, you’ll need to write the solution as an arcfunction.

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Trig equations have one important difference from other types of
equations.
**Trig functions are periodic**,
meaning that they repeat their values over and over.
Therefore a trig equation has
an **infinite number of solutions** if it has any.

Think about an
equation like sin *u* = 1.
π/2 is a solution, but the sine function repeats its values every
2π. Therefore π/2±2π,
π/2±4π, and so on are equally good solutions. To show this,
write the solution as *u* = π/2 + 2π*n*,
where *n* is understood to be any integer, positive, negative, or zero.
(The tangent and cotangent functions repeat all their values every π
radians, so the solution to tan *v* = 1 is
*v* = π/4 + π*n*, not +2π*n*.)

**Example A** continues. Recapping what was done so far,

cos(4*A*) − sin(2*A*) = 0 ⇒

2sin²(2*A*) + sin(2*A*) − 1 = 0 ⇒

sin(2*A*) = −1 or sin(2*A*) = 1/2

The sine of 3π/2 is −1, so the first possibility reduces to
2*A* = 3π/2. But remember that the sine function is periodic,
so write

sin(2*A*) = −1 ⇒
2*A* = 3π/2 + 2π*n*.

For the second possibility, sin(2*A*) = 1/2, there are two
solutions, because sin(π/6) and sin(5π/6) both equal 1/2, and again we
add 2π*n* to the angle to account for all solutions:

sin(2*A*) = 1/2 ⇒ 2*A* = π/6 + 2π*n* or 5π/6 + 2π*n*

Combining these, here are the three solutions for the original equation:

2*A* = 3π/2 + 2π*n* or π/6 + 2π*n* or 5π/6 + 2π*n*

**Example B** continues. Recapping what was done so far,

3tan²(*B*/2) − 1 = 0 ⇒

tan(*B*/2) = ±√3/3

What angle has a tangent value of √3/3? the angle π/6. And where does the tangent have a value of −√3/3? at the angle 5π/6. This gives the solutions

*B*/2 = π/6 + π*n* or 5π/6 + π*n*

Remember that the tangent and cotangent have period π and not 2π.

**Example C**:

Of course, you don’t always luck out with nice angles. Take a look at this equation:

sec(3*C*) = 2.5

What are the possible values of the angle 3*C*?
It’s hard to work with the secant function, but
1/sec(3*C*) = cos(3*C*) so rewrite the equation as

1/sec(3*C*) = 1/2.5

cos(3*C*) = 0.4

For what angles is that true? We write arccos(0.4) to mean
the angle in quadrant I that has a cosine equal to 0.4. (Some books
write cos^{−1}(0.4) instead of arccos(0.4). I prefer the arccos
notation because the superscript −1 makes many students think of
1/cos(0.4), which has a different meaning entirely.)

So initially we would write 3*C* = arccos(0.4) + 2π*n*.
But that’s
not the whole story: any angle in quadrant I has a reflection in
quadrant IV with the same cosine value,
so we need to account for both angles:

3*C* = arccos(0.4) + 2π*n* or 2π−arccos(0.4) + 2π*n*

Note that the base angle is always nonnegative and less than 2π:
2π−arccos(0.4) + 2π*n*, not simply
−arccos(0.4) + 2π*n*.
This is necessary to make step 5 come out right.

Now it’s time to abandon angular
thinking and go for the variable. As you will see, it is very
important to do this step *after*
step 3, not before.
2π*n* or π*n* must be added to the angle, not the variable,
to reflect the period of the trig function.

**Example A** continues. Recapping what was done so far,

cos(4*A*) − sin(2*A*) = 0 ⇒

2sin²(2*A*) + sin(2*A*) − 1 = 0 ⇒

sin(2*A*) = −1 or sin(2*A*) = 1/2 ⇒

2*A* = 3π/2 + 2π*n* or π/6 + 2π*n* or 5π/6 + 2π*n*

Now divide both sides by 2 to solve for
the variable, *A*:

*A* = 3π/4 + π*n* or π/12 + π*n* or 5π/12 + π*n*

Be sure to divide the entire equation, so that the 2π*n* becomes π*n*.

Why is the order of steps so important?
The 2πn came in because the sine function has a period of 2π:
if you take an angle and add 2π to it, it looks like the same
angle and all six of its function values are unchanged. But now we’re no
longer dealing with the *angle* 2*A*, we’re dealing with the
*variable* *A*. In this equation, we say that adding π to any
solution for *A* will give another solution for *A*.

For instance, set *n*=1 and obtain solutions *A* = 7π/4, 13π/12,
or 17π/12.
True, sin(7π/4) doesn’t equal −1. But the equation was
sin(2*A*) = −1, not sin(*A*) = −1. If you
substitute *A* = 7π/4 in sin(2*A*), you get sin(2·7π/4) =
sin(7π/2) which does equal −1. Always pay attention to whether you’re
dealing with the angle or the variable.

**Example B** continues. Recapping what was done so far,

3tan²(*B*/2) − 1 = 0 ⇒

tan(*B*/2) = ±√3/3 ⇒

*B*/2 = π/6 + π*n* or 5π/6 + π*n*

Multiplying both sides by 2 gives

*B* = π/3 + 2π*n* or 5π/3 + 2π*n*

Once again, the angle was *B*/2 and had a period of π; the variable
is *B* and has a period of 2π.

**Example C** continues. Recapping what was done so far,

sec(3*C*) = 2.5 ⇒

3*C* = arccos(0.4) + 2π*n* or 2π−arccos(0.4) + 2π*n*

Divide both sides by 3:

*C* = (1/3)arccos(0.4) + (2π/3)*n* or (2π/3)−(1/3)arccos(0.4) + (2π/3)*n*

It’s a matter of taste whether to combine terms in that second solution:

*C* = (1/3)arccos(0.4) + (2π/3)*n* or −(1/3)arccos(0.4) + (2*n*+1)π/3

Does the problem specify a solution interval for the variable?
Sometimes this is done in interval notation, like [0;2π); other times it’s
done as an inequality, 0 <= *x* < 2π. If no
restriction is given, you should give all real solutions as shown in
step 4.

But if solutions are restricted to a particular interval, you have a bit more work to do after solving for the variable.

**Example A** continues. Recapping what was done so far,

cos(4*A*) − sin(2*A*) = 0 ⇒

2sin²(2*A*) + sin(2*A*) − 1 = 0 ⇒

sin(2*A*) = −1 or sin(2*A*) = 1/2 ⇒

2*A* = 3π/2 + 2π*n* or π/6 + 2π*n* or 5π/6 + 2π*n* ⇒

*A* = 3π/4 + π*n* or π/12 + π*n* or 5π/12 + π*n*

Now suppose that only solutions on the interval [0;2π) were wanted.

The *general* solutions have a period of π (from +π*n*),
and therefore there will be two cycles between 0 and 2π:

n |
A = 3π/4 + πn |
A = π/12 + πn |
A = 5π/12 + πn |
---|---|---|---|

0 | 3π/4 | π/12 | 5π/12 |

1 | 7π/4 | 13π/12 | 17π/12 |

The solutions for *n* = 2 are all larger than the 2π boundary
of the interval. There are six solutions to the equation for *A* within
the interval [0;2π). In order, they are π/12, 5π/12, 3π/4,
13π/12, 17π/12, and 7π/4.

**Example B** continues. Recapping what was done so far,

3tan²(*B*/2) − 1 = 0 ⇒

tan(*B*/2) = ±√3/3 ⇒

*B*/2 = π/6 + π*n* or
5π/6 + π*n* ⇒

*B* = π/3 + 2π*n* or 5π/3 + 2π*n*

Suppose that the problem specified solutions between 0 and π/2.
As you can see, even with *n* = 0 there is just one solution
within the limits [0;π/2). Answer: π/3 only.

**Example C** continues. Recapping what was done so far,

sec(3*C*) = 2.5 ⇒

3*C* = arccos(0.4) + 2π*n* or 2π−arccos(0.4) + 2π*n* ⇒

*C* = (1/3)arccos(0.4) + (2π/3)*n* or −(1/3)arccos(0.4) + (2*n*+1)π/3

Suppose again that limits of *C* in [0;2π) are stated.
The period of the variable is 2π/3, so there are three cycles
(*n*=0,1,2) between 0 and 2π.
Here are the values, with decimal approximations:

n |
C = (1/3)arccos(0.4) + (2π/3)n |
C = −(1/3)arccos(0.4) + (2n+1)π/3 |
---|---|---|

0 | (1/3)arccos(0.4) ≈ 0.3864 | −(1/3)arccos(0.4) + π/3 ≈ 0.6608 |

1 | (1/3)arccos(0.4) + 2π/3 ≈ 2.4808 | −(1/3)arccos(0.4) + 3π/3 ≈ 2.7552 |

2 | (1/3)arccos(0.4) + 4π/3 ≈ 4.5752 | −(1/3)arccos(0.4) + 5π/3 ≈ 4.8496 |

Since 2π is about 6.28, there are six solutions within the stated limits.
*C* = about 0.3864, 0.6608, 2.4808, 2.7552, 4.5752, and 4.8496.

**Example D**:

Solve ½sin(2*D*) − ½ = 0 for *D* in [0;2π).

Solution: There is only one function (sin) of one angle (2*D*),
so step 1 is complete. Don’t “simplify”
½sin(2*D*) to sin(*D*): that is not a valid
operation. And don’t
convert sin(2*D*) to 2 sin(*D*) cos(*D*); while that
*is* mathematically valid, it makes the equation more
complicated, not simpler.

Step 2: solve for the function.

½sin(2*D*) = ½

sin(2*D*) = 1

Step 3: find the angle.

2*D* = π/2 + 2π*n*

Step 4: solve for the variable.

n | D = π/4 + πn |
---|---|

0 | π/4 |

1 | 5π/4 |

*D* = π/4 + π*n*

Step 5: apply any restrictions. The period of the variable is π, which means there will be two cycles in the interval [0;2π).

Answer: *D* = π/4, 5π/4.

**Example E**:

Find all solutions for sin²(*E*/2) − cos²(*E*/2) = 1.

Solution: In step 1, you use
identities to get the
equation in term of one function of one angle.
As often happens in
trig, you have a choice of which identity you want to use.
You could replace cos²(*E*/2) with
1−sin²(*E*/2), or you could use the
double-angle formula
cos(2*u*) = cos²(*u*) − sin²(*u*)
with *u* = *E*/2.
That second approach leads to a simpler equation:

sin²(*E*/2) − cos²(*E*/2) = 1

cos²(*E*/2) − sin²(*E*/2) = −1

cos(2*E*/2) = −1

cos(*E*) = −1

Step 2 is already done.

Step 3: write down the angle:

*E* = π + 2π*n* = (2*n*+1)π

Step 4 is already done, since the angle is the variable.

Step 5 is easily done: the problem
specifically asks for all values of *E*.

Answer:
*E* =(2*n*+1)π for any integer *n*.

**28 Oct 2020**: Updated the Summary to explain the terms “numerically” and “closed form”.Showed an intermediate step in a “straightforward” solution.

Add the possibility of angles being multiples of half of a special angle.

Made a few additional tweaks for clarity.

Converted page from HTML 4.01 to HTML5, and aitalicized the variable names.

**17 Aug 2015**: Moved from OakRoadSystems.com to BrownMath.com.- (intervening changes suppressed)
**20 Apr 2002**: First publication.

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