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Solving Trigonometric Equations

Copyright © 2002–2015 by Stan Brown

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. But a great many can be solved in closed form, 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:

  1. Put the equation in terms of one function of one angle.
  2. Write the equation as one trig function of an angle equals a constant.
  3. Write down the possible value(s) for the angle.
  4. If necessary, solve for the variable.
  5. 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.

Step 1. Get one function of one angle.

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(4A) − sin(2A) = 0

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

As you see, that equation involves two functions (sine and cosine) of two angles (4A and 2A). 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(2u) can be expressed in terms of sin(u). Specifically, cos(2u) = 1 − 2sin²(u).

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

cos(2u) = 1 − 2sin²(u)

cos(2×2A) = 1 − 2sin²(2A)

cos(4A) = 1 − 2sin²(2A)

That transforms the original equation to

1 − 2sin²(2A) − sin(2A) = 0

which can be rewritten in standard form as

2sin²(2A) + sin(2A) − 1 = 0

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

Step 2. Solve for the value(s) of a trig function.

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(4A) − sin(2A) = 0  ⇒

2sin²(2A) + sin(2A) − 1 = 0

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

2y² + y − 1 = 0, where y = sin(2A)

It can be factored in a straightforward way:

(sin(2A) + 1) (2sin(2A) − 1) = 0

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

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

sin(2A) = −1       or       sin(2A) = 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)  or  ±√(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.

Step 3. Solve for the angle.

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. Otherwise you must write the solution as an arcfunction.

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 + 2πn, 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 + πn, not +2πn.)

Example A continues. Recapping what was done so far,

cos(4A) − sin(2A) = 0  ⇒

2sin²(2A) + sin(2A) − 1 = 0 ⇒

sin(2A) = −1       or       sin(2A) = 1/2

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

sin(2A) = −1 ⇒ 2A = 3π/2 + 2πn.

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

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

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

2A = 3π/2 + 2πn    or    π/6 + 2πn     or    5π/6 + 2πn

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 + πn    or     5π/6 + πn

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(3C) = 2.5

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

1/sec(3C) = 1/2.5

cos(3C) = 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 3C = arccos(0.4) + 2πn. 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:

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

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

Step 4. Solve for the variable.

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. 2πn or πn 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(4A) − sin(2A) = 0  ⇒

2sin²(2A) + sin(2A) − 1 = 0 ⇒

sin(2A) = −1   or   sin(2A) = 1/2 ⇒

2A = 3π/2 + 2πn   or   π/6 + 2πn    or   5π/6 + 2πn

Now divide both sides by 2:

A = 3π/4 + πn   or   π/12 + πn   or   5π/12 + πn

Be sure to divide the entire equation, so that the 2πn becomes πn.

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 2A, 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(2A) = −1, not sin(A) = −1. If you substitute A = 7π/4 in sin(2A), 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 + πn   or   5π/6 + πn

Multiplying both sides by 2 gives

B = π/3 + 2πn   or   5π/3 + 2πn

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(3C) = 2.5 ⇒

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

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) + (2n+1)*π/3

Step 5. Apply any restrictions.

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(4A) − sin(2A) = 0  ⇒

2sin²(2A) + sin(2A) − 1 = 0 ⇒

sin(2A) = −1   or   sin(2A) = 1/2 ⇒

2A = 3π/2 + 2πn   or   π/6 + 2πn    or   5π/6 + 2πn ⇒

A = 3π/4 + πn   or   π/12 + πn   or   5π/12 + πn

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

The general solutions have a period of π (from +πn), 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/4π/125π/12
1 7π/413π/1217π/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 + πn   or   5π/6 + πn ⇒

B = π/3 + 2πn   or   5π/3 + 2πn

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(3C) = 2.5 ⇒

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

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

More Examples

Example D:

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

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

Step 2: solve for the function.

½sin(2D) = ½

sin(2D) = 1

Step 3: find the angle.

2D = π/2 + 2πn

Step 4: solve for the variable.

nD = π/4 + πn

D = π/4 + πn

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, use identities to get the equation in term of one function of one angle. Here you have a choice. You could replace cos²(E/2) with 1−sin²(E/2), or you could use the double-angle formula cos(2u) = 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 + 2πn = (2n+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 =(2n+1)π for any integer n.

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