SOLVING COMPLEX TRIGONOMETRIC EQUATIONS

Example 1:

Solve 2sin² x 3sin x + 1 = 0 for x on the interval . Express the solutions in degrees.

Step 1:

This equation is "quadratic" in form. That is, one variable (sin x) is raised to the second power and the other, same variable, is raised to the first power.

Nothing needs to be done because the equation only contains sine ratios.

Step 2:

We are going to attempt factoring. To make factoring easier, we will make a little substitution. That is, we let sin x equal some letter, say k.

Then we rewrite 2sin² x 3sin x + 1 as 2k²3k + 1. It can be factored as (2k 1)(k 1).

Then it should be easy to see that we can write

2sin² x 3sin x + 1 = (2sin x 1)(sin x 1) = 0

Using the Zero Product Principle, we will now set both factors equal to 0 and then solve for x.

(1) 2sin x 1 = 0 then sin x =

(2) sin x 1 = 0 then sin x = 1

We now have two simple trigonometric equations. We must try to find solutions for both. The solutions of cos2x + 3sin x 2 = 0 will be a combination of both.

Steps 3 through 6:

(1) Let's use the calculator to solve sin x = . See Lesson 12 - Solving Simple Trigonometric Equations.

Using the arcsine, we find x = 30^o, which indicates a reference angle of 30^o.

Next, we will use All Students Take Calculus to find the quadrants in which the sine ratio is positive. It is best to draw a picture!

We need to find angles located in QI and QII that have a reference angle of 30^o. We find the solutions for angle x on the interval to be as follows:

Solution in QI:

x₁ = 30^o

Solution in QII:

x₂ = 180^o30^o = 150^o

(2) Let's use the calculator to solve sin x = 1. See Lesson 12 - Solving Simple Trigonometric Equations.

With the calculator in degrees and using the arcsine, we find x = 90^o, which indicates a Quadrantal Angle, therefore, there is no reference angle.

Next, we use the following picture:

We find one solution for angle x on the interval to be 90^o.

In summary, the solutions of 2sin² x 3sin x + 1 = 0 on the interval are 30^o, 90^o, and 150^o.

Example 2:

Solve 2cos²x + 3sin x = 0 for x on the interval .  Express the solutions in degrees.

Step 1:

This equation is "quadratic" in form. However, it contains sine and cosine ratios. We must change the equation so that it contains either all cosines or all sines.

Remembering the Pythagorean Identity cos²x = 1 sin² x, we can change the given equation to the following:

2(1 sin² x) + 3sin x = 0          

Using the Distributive Property:

2 2sin² x + 3sin x = 0

Writing in descending powers of sin x:

2sin² x + 3sin x + 2 = 0

Finally, we will multiply both sides by 1 to get 2sin² x 3sin x 2 = 0.

Step 2:

We are going to attempt factoring. To make factoring easier, we will now make a little substitution. That is, we let sin x equal some letter, say k.

Then we rewrite 2sin² x 3sin x 2 as 2k²3k 2. It can be factored as (2k + 1)(k 2).

Then it should be easy to see that we can write

2sin² x 3sin x 2 = (2sin x + 1)(sin x 2) = 0

Using the Zero Product Principle, we will now set both factors equal to 0 and then solve for x.

(1) 2sin x + 1 = 0 then sin x =

(2) sin x 2 = 0 then sin x = 2

We now have two simple trigonometric equations. We must try to find solutions for both. The solutions of 2cos²x + 3sin x = 0 will be a combination of both.

Steps 3 through 6:

(1) Let's use the calculator to solve sin x = . See Lesson 12 - Solving Simple Trigonometric Equations.

With the calculator in degrees and using the arcsine, we find x = 30^o, which indicates a reference angle of 30^o.

Next, we will use All Students Take Calculus to find the quadrants in which the sine ratio is negative. It is best to draw a picture!

We need to find angles located in QIII and QIV that have a reference angle of 30^o. We find the solutions for the angle x on the interval to be as follows:

Solution in QIII:

x₁ = 180^o + 30^o = 210^o

Solution in QIV:

x₂ = 360^o 30^o = 330^o

(2) Let's use the calculator to solve sin x = 2. See Lesson 12 - Solving Simple Trigonometric Equations.

Using the arcsine, we get a Domain Error!

Why?

Please note that sin x is never greater than 1 (remember its graph?), therefore, sin x = 2 is undefined, therefore, has no solution..

In summary, the solutions of 2cos²x + 3sin x = 0 on the interval are 210^o and 330^o.

Solve sin²x 2cos x 2 = 0 for x on the interval . Express the solutions in degrees.

Step 1:

This equation is "quadratic" in form. However, it contains sine and cosine ratios. We must change the equation so that it contains either all cosines or all sines.

Remembering the Pythagorean Identity sin²x = 1 cos² x, we can change the given equation to the following:

(1 cos² x) 2cos x 2 = 0

Writing in descending powers of cos x:

cos² x 2cos x 1 = 0

Finally, we will multiply both sides by 1 to get cos² x + 2cos x + 1 = 0.

Step 2:

We are going to attempt factoring. To make factoring easier, we will make a little substitution. That is, we let cos x equal some letter, say k.

Then we rewrite cos² x + 2cos x + 1 as k² + 2k + 1. It can be factored as (k + 1)(k + 1).

Then it should be easy to see that we can write

cos² x + 2cos x + 1 = (cos x + 1)(cos x + 1) = 0

We have two identical factors. We will use the Zero Product Principle to set one equal to 0 and then solve for x.

cos x + 1 = 0 then cos x = 1

We now have one simple trigonometric equation. We must try to find its solutions. They will be the solutions of sin²x 2cos x 2 = 0 as well.

Steps 3 through 6:

Let's use the calculator to solve cos x = 1. See Lesson 12 - Solving Simple Trigonometric Equations.

With the calculator in degrees and using the arccosine, we find x = 180^o, which indicates a Quadrantal Angle, therefore, there is no reference angle.

Next, we use the following picture:

In summary, there is only one solution of sin²x 2cos x 2 = 0 on the interval , namely x = 180^o.

Example 4:

Solve cot x = 4 sin² x cot x for x on the interval .  Express the solutions in radians.

Step 1:

This equation contains sine and cotangent ratios with one of the cotangent ratios being connected to the squared sine ratio via multiplication. Therefore, we will NOT attempt to change the equation so that it contains either all sines or all cotangents.

Step 2:

We are going to solve this equation by factoring out the greatest common factor which seems to be cot x.

cot x = 4 sin² x cot x

0 = 4 sin² x cot x cot x

0 = cot x (4 sin² x 1)

Using the Zero Product Principle, we will now set both factors equal to 0 and then solve for x.

(1) cot x = 0

(2) 4 sin² x 1 = 0

then and

We now have three simple trigonometric equations. We must try to find solutions for all of them. The solutions of cot x = 4 sin² x cot x will be a combination of them all.

Steps 3 through 6:

(1) Finding the angle x for cot x = 0 with the calculator is cumbersome.

Hopefully, we remember that the cotangent ratio is only 0 when the angle is quadrantal. We also hopefully remember that .

The sine ratio cannot equal 0 otherwise the cotangent ratio is undefined. Let's look at the following picture.

At which angles is the sine ratio NOT equal to 0? We find two solution for angle x on the interval , namely /2 and 3/2.

(2) Let's use the calculator to solve sin x = and sin x = . See Lesson 12 - Solving Simple Trigonometric Equations.

With the calculator in degrees and using the arcsine, we find x = 30^o and x = 30^o, which indicates a reference angle of 30^o in both cases.

Next, we will use All Students Take Calculus to find the quadrants in which the sine ratio is both positive and negative. It is best to draw a picture!

   and       

We need to find angles located in QI, QII, QIII, and QIV that have a reference angle of 30^o. We find the solutions for angle x on the interval to be as follows:

Solution in QI:

x₁ = 30^o /6

Solution in QII:

x₂ = 180^o 30^o = 150^o 5/6

Solution in QIII:

x₂ = 180^o + 30^o = 210^o 7/6

Solution in QIV:

x₂ = 360^o 30^o = 330^o 11/6

In summary, the solutions of cot x = 4 sin² x cot x on the interval are

/2, 3/2, /6 , 5/6, 7/6, and 11/6.

Example 5:

Solve for x on the interval .  Express the solutions in degrees.

Step 1:

This equation is "quadratic" in form. However, it only contains a squared cosine ratio.

Step 2:

We are going to solve this equation for cos x by using the Square Root Property.

First, we must change the leading coefficient of the square trigonometric term to 1 by dividing both sides of the equation by 2.

We can now apply the Square Root Property to . Subsequently we get

We now have two simple trigonometric equations, namely

and

We must try to find solutions for both. The solutions of will be a combination of both.

Steps 3 through 6:

Let's use the calculator to solve and . See Lesson 12 - Solving Simple Trigonometric Equations.

With the calculator in degrees and using the arccosine, we find x = 45^o and x = 135^o, which indicates a reference angle of 45^o in both cases.

Next, we will use All Students Take Calculus to find the quadrants in which the cosine ratio is both positive and negative. It is best to draw a picture!

   and       

We need to find angles located in QI, QII, QIII, and QIV that have a reference angle of 45^o. We find the solutions for the angle x on the interval to be as follows:

Solution in QI:

x₁ = 45^o

Solution in QII:

x₂ = 180^o 45^o = 135^o

Solution in QIII:

x₃ = 180^o + 45^o = 225^o

Solution in QIV:

x₄ = 360^o 45^o = 315^o

In summary, the solutions of on the interval are 45^o, 135^o, 225^o, and 315^o.

Example 6: (Trigonometric equations involving multiples of an angle)

Solve for x on the interval .  Express the solutions in degrees.

Here we we have an equation were bx = which is the same as x.

Please note that we will work with the angle through Step 5. Only in Step 6 will we find the solution(s) for angle x.

Step 1:

The solution interval for angle x is .

The solution interval for angle is half of this, namely . 

Steps 2 through 5:

See Lesson 12 - Solving Simple Trigonometric Equations.

(2) Isolate the trigonometric expression on one side:

(3) We are going to use the concept of Inverse Trigonometric Functions to solve for the angle using the calculator in degree mode.

Please note that you cannot at this point multiply both sides of the equation by 2 to get a solution for angle x!!!

(4) The reference angle for angle equals 45^o.

(5) The value of the cosine ratio in (2) is positive and we have a reference angle.

We will use All Students Take Calculus to find the quadrants in which the cosine ratio is positive. It is best to draw a picture!

Finally, we are going to use the reference angle from Step 3.  We need to find angles located in QI and QIV that have a reference angle of 45^o. 

Therefore, the solutions for angle on the interval are as follows:

= 45^o

and = 360^o45^o = 315^o

Step 6:

Since we are asked to solve for angle x, we must now multiply both sides of the above equations by 2 to get

In summary, the solutions of on the interval are 90^o and 630^o.

Example 7: (Trigonometric equations involving multiples of an angle)

Solve for x on the interval .  Express the solutions in EXACT radians.

Here we we have an equation were bx = 4x.

Please note that we will work with the angle 4x through Step 5. Only in Step 6 will we find the solution(s) for angle x.

Step 1:

The solution interval for angle x is .

The solution interval for angle 4x is four times as large as this, namely . 

Steps 2 through 5:

See Lesson 12 - Solving Simple Trigonometric Equations.

(2) Isolate the trigonometric expression on one side:

(3) We are going to use the concept of Inverse Trigonometric Functions to solve for the angle 4x using the calculator. 

Let's work with degrees! Using the calculator in degree mode, we find

Please note that you cannot at this point divide both sides of the equation by 4 !!!

(4) The reference angle of angle 4x equals 30^o.

(5) The value of the sine ratio in (2) is positive and we have a reference angle.

Qe will use All Students Take Calculus to find the quadrants in which the sine ratio is positive. It is best to draw a picture!

Finally, we are going to use the reference angle from Step 3.  Now, remember the solution interval for angle 4x is in degrees.

However, we are going to find angles in the solution interval first. They must be in QI and QII and have a reference angle of 30^o. 

They are

4x₁ = 30^o

4x₂ = 180^o 30^o = 150^o

Since we are supposed to find solutions for angle 4x on the interval , we must add the following angles, which are coterminal to the two we already found:

4x₃ = 30^o + 360^o = 390^o

4x₄ = 150^o + 360^o = 510^o

Let's go ahead and express these four solutions in radians as follows:

                                                       

Step 6:

Since we are asked to solve for the angle x, we must now divide both sides of the above equations by 4 to get

                           

In summary, the solutions of on the interval are

                            .