Solving second degree trigonometric equations

Trigonometric Equations

Put simply, trigonometric equations are just equations that feature the trigonometric ratios such as sine and cosine on the variable “x”. Because of the presence of these trigonometric functions, solving these equations becomes a little more difficult. But, before we get into solving these trig equations, let’s make sure we understand what 2nd degree trig equations are! Below are a couple examples of trig equations:

sin2x+sinx=4\sin ^2x+\sin x =4

cos2y4cosy+3=0\cos ^2y-4 \cos y +3 =0

Notice how the above equations feature our familiar polynomial format but with the addition of the trigonometric ratios sine and cosine. Now that we have an idea of what trig equations look like, let’s look at how to solve trig equations!

Solving Trigonometric Equations

In order to solve the trigonometry problems we’ll be looking at in this article, it is first important to make sure you are comfortable with basic, non-trigonometric functions and relations, as well as factoring polynomials. To refresh your memory, videos on multiplying functions and factoring trinomials will be of great help!

With those concepts in mind, let’s get to solving trig equations. The are many different methods to solving trig problems, so depending where you look, you may get many different answers! This article will give an in-depth overview of the 2 main methods we can use for solving trigs.

Method 1:

Replace sin x or cos x with another variable, solve, and then replace that variable with sin x or cos x.

Method 2:

Solve without doing any replacement. This method requires that you have a good grasp of trigonometric functions and are comfortable simplifying expressions with them!

Now, it’s quite difficult to describe these methods without an example to show, so let’s practice these methods with some actual problems. After all, the best way to learn how to solve second-degree trig equations is to do some practice problems!

Example 1:

Solve the following equation:

6sin2x+3sinx=06 \sin ^2 x + 3 \sin x = 0

for0x2πfor\;0 \leq x \leq 2 \pi

This question is asking us to find solutions for x within the range of 0 to 2pi. In this example, we’ll first use Method 1 to solve it.

Step 1: Replace sinx with another variable

Let w=sinxw = \sin x


Step 2: Factor the newly-formed polynomial



Step 3: Solve for x

Hopefully you remember how to solve factored polynomials. In this case, w = 0 and w = -1/2

Step 4: Replace w with sinx and change solutions

Now that we have solved the equation for w, we have to remember to replace w back with sin x and adjust our solutions to accommodate this.

Therefore: sin x = 0 and sin x = -1/2

Using our calculator or our knowledge of the ever-important unit circle, we arrive at the following solutions: x = 0 and x = -pi/6

For practice solving these kinds of calculations, check out the lesson on finding trig values.

Step 5: Find solutions within entire specified range

Though we have found solutions for our equation, remember we were asked to find all of the solutions within the range of 0x2π0 \leq x \leq 2 \pi! Because of the repeating nature of sine and cosine functions, we’re likely to have several values for x that can solve this equation. Furthermore, -pi/6 does not solve this equation, as it is not within the range.

Again, as with most things in mathematics, there are several ways to solve for the other values of x. The easiest way is to quickly sketch out a sine graph from 0 to 2pi and see when pi is equal to 0 and -1/2.

Taking a look at our image above, it is easy to spot where sinx = 0, giving us the x-values of 0, pi, and 2pi. Now looking for sinx = -1/2, we know from our initial solution of –pi/6 that sinx = -1/2 for some multiples of pi/6. Looking at our graph, it is clear that sinx = -1/2 for two values between pi and 2pi, those values being pi + pi/6 and 2pi – pi/6. Therefore, this gives us the x-values of 7pi/6 and 11pi/6.

Therefore, our final answer is: x = 0, pi, 7pi/6, 11pi/6, and 2pi

Another method to solving for the above extra solutions involves knowledge of reference angle and the ASTC Rule, which are slightly more complicated. Be sure to watch the videos from those lessons to learn more!

Example 2:

Solve the following equation:

2cos2x+3 cosx+1=02 \cos^2 x + 3\ cos x +1=0

For this second, and last, example, we’ll be using Method 2 that was described earlier. That means we won’t be doing any replacement, and will be solving the trig equation as-is.

Step 1: Factor and simplify the equation

(2cosx+1)(cosx+1)=0(2 \cos x+1) ( \cos x +1)=0

Step 2: Solve

Now that we’ve factored our equation, it’s easy to solve and determine that:

cosx = -1/2 and cosx = -1

Solving for x gives us the solutions: x = 2pi/3 and pi

Step 3: Find solutions within entire specified range

In this example, contrary to the first one, no specified range was given. Therefore, we must give solutions that cater to the ever-repeating nature of the cosine function. Also, we need to consider that there may be other values for x that our calculators didn’t spit out. Again, the best way to figure all of this out is to take a look at the graph of cosx.

Taking a look at the graph above, again, it’s easy to spot when cosx = -1. Cosx will equal -1 at pi, repeating every “nth” (where n = an integer) full cycle (2pi) of the wave, thus x = pi + n2pi. For cosx=-1/2, the solution is similar to how we found our solutions in Example 1. Also, we need to remember the repeating nature as we did for solutions for cosx = -1 for every full cycle (2pi). Thus, the function will equal -1/2 when x = 2pi/3 = n2pi and when x = 4pi/3 + n2pi.

Therefore, our final answer is: thus x = pi + n2pi, x = 2pi/3 = n2pi, and x = 4pi/3 + n2pi

All that’s all there is to it! For more studies into trigonometric equations, and how they relate to taking derivatives, check out this lesson on derivatives of trig functions.

Solving second degree trigonometric equations


  • 1.
    Solve the following trigonometric equations:
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Solving second degree trigonometric equations

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