SOLVING SIMPLE TRIGONOMETRIC EQUATIONS

Example 1:

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

Step 1:

The interval in which we are supposed to find the solutions is . We will now isolate the trigonometric ratio by dividing both sides by 2 to get

We notice that solution angles must be in QI and QII.

Step 2:

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

Helpful Hint: It's easier to work in degrees and then change the solutions back to radians! In this case, we will change the solution interval to degrees, namely .

Step 3:

Given the solution in Step 2, we will find its reference angle. Since angle 45^o is a QI angle, its reference angle equals 45^o.

Step 4:  

The value of the sine ratio in Step 1 is positive and we have a reference angle. 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 solution angles located in QI and QII with a reference angle of 45^o. We find the solutions for angle x in the interval to be as follows.

Solution in QI:

x₁ = 45^o

Solution in QII:

x₂ = 180^o 45^o = 135^o

Since we are supposed to express the solutions in terms of radians we get

x₁ = /4

and x₂ = 3/4

Example 2:

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

Step 1:

The interval in which we are supposed to find the solutions is . We will now isolate the trigonometric ratio by dividing both sides by 2 to get

We notice that solution angles must be in QIII and QIV.

Step 2:

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

Helpful Hint: It's easier to work in degrees and then change the solutions back to radians! In this case, we will change the solution interval to degrees, namely .

Step 3:

Given the solution in Step 2, we will find its reference angle. We know that the reference angle of a negative angle is equal to that of its positive counterpart. 

Since 45^o is a QI angle, its reference angle equals 45^o. Therefore, the reference angle of 45^o is 45^o as well.

Step 4:

The value of the sine ratio in Step 1 is negative and we have a reference angle. 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 solution angles located in QIII and QIV with a reference angle of 45^o. We find the solutions for angle x in the interval to be as follows.

Solution in QIII:

x₁ = 180^o + 45^o = 225^o

Solution in QIV:

x₂ = 360^o 45^o = 315^o

Since we are supposed to express the solutions in terms of radians we get

x₁ = 5/4

and x₂ = 7/4

Example 3:

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

Step 1:

The interval in which we are supposed to find the solutions is . The trigonometric ratio is already isolated on one side of the equal sign. 

We notice that solution angles must be in QI and QIII.

Step 2:

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

Helpful Hint: It's easier to work in degrees and then change the solutions back to radians! In this case, we will change the solution interval to degrees, namely .

Step 3:

Given the solution in Step 2, we will find its reference angle. Since angle 60^o is a QI angle, its reference angle equals 60^o.

Step 4:  

The value of the tangent ratio in Step 1 is positive and we have a reference angle. Next, we will use All Students Take Calculus to find the quadrants in which the tangent ratio is positive. It is best to draw a picture!

 

We need to find solution angles located in QI and QIII with a reference angle of 60^o. We find the solutions for angle x in the interval to be as follows.

Solution in QI:

x₁ = 60^o

Solution in QIII:

x₂ = 180^o + 60^o = 240^o

Since we are supposed to express the solutions in terms of radians we get

x₁ = /3

and x₂ = 4/3

Example 4:

We are going to solve again, but now we only want solutions for x in the interval .  Express the solutions in EXACT radians.

Step 1:

The interval in which we are supposed to find the solutions is . The trigonometric ratio is already isolated on one side of the equal sign. 

We notice that solution angles must be in QI and QIII.

Step 2:

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

Helpful Hint: It's easier to work in degrees and then change the solutions back to radians! In this case, we will change the solution interval to degrees, namely [180^o,180^o).

Step 3:

Given the solution in Step 2, we will find its reference angle. Since angle 60^o is a QI angle, its reference angle equals 60^o.

Step 4:

The value of the tangent ratio in Step 1 is positive and we have a reference angle. Next, we will use All Students Take Calculus to find the quadrants in which the tangent ratio is positive. It is best to draw a picture!

 

We need to find solution angles located in QI and QIII with a reference angle of 60^o.

Here Step 4 is a bit more complicated because we are asked to find solutions in the interval [180^o,180^o). Drawing a picture is always a good idea.

Since we determined that the solutions are in QI and QIII, let's draw the reference angles in these quadrants AND the solution interval [180^o,180^o).

We now find the solutions for angle x in the interval [180^o, 180^o] with the help of the picture above.

Solution in QI:

x₁ = 60^o

Solution in QIII:

x₂ = 180^o + 60^o = 120^o

Since we are supposed to express the solutions in terms of radians we get

x₁ = /3

and x₂ = 2/3

Example 5:

Solve cos x = 0.858 for x in the interval [265^o, 90^o] in degrees.  Round all calculations to 2 decimal places.

Step 1:

The interval in which we are supposed to find the solutions is [265^o, 90^o]. The trigonometric ratio is already isolated on one side of the equal sign. 

cos x = 0.858

We notice that solution angles must be in QII and QIII.

Step 2:

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

cos x = 0.858

x = cos^-1 (0.858)

x 149.09^o

Step 3:

Given the solution in Step 2, we will find its reference angle. Since 149.09^o is a QII angle, its reference angle is 180^o 149.09^o = 30.91^o.

Step 4:

The value of the cosine ratio in Step 1 is negative and we have a reference angle. Next, we will use All Students Take Calculus to find the quadrants in which the cosine ratio is negative. It is best to draw a picture!

We need to find solution angles located in QII and QIII with a reference angle of 30.91^o.

Here Step 4 it a bit more complicated because we are asked to find solutions in the interval [265^o, 90^o]. Drawing a picture is always a good idea.

Since we determined that the solutions are in QII and QIII, let's draw the reference angles in these quadrants and the solution interval [265^o, 90^o].

We now find the solutions for angle x in the interval [265^o, 90^o] with the help of the picture above.

Solution in QII:

x₁ = [ 30.91^o + 180^o] = 210.91^o

Solution in QIII:

x₂ = 180^o + 30.91^o = 149.09^o

Example 6:

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

Step 1:

The interval in which we are supposed to find the solutions is . We will isolate the trigonometric ratio by dividing both sides by 1 to get

Step 2:

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

Helpful Hint: It's easier to work in degrees and then change the solutions back to radians! In this case, we will change the solution interval to degrees, namely .

Step 3:

Given the solution in Step 2, we will find its reference angle.We know that the reference angle of a negative angle is equal to that of its positive counterpart. 

Since 60^o is a QI angle, its reference angle equals 60^o. Therefore, the reference angle of 60^o is 60^o as well.

Step 4:

The value of the tangent ratio in Step 1 is negative and we have a reference angle. Next, we will use All Students Take Calculus to find the quadrants in which the tangent ratio is negative. It is best to draw a picture!

We need to find solution angles located in QII and QIV with a reference angle of 60^o. We find the solutions for angle x in the interval to be as follows:

Solution in QII:

x₁ = 180^o 60^o = 120^o

Solution in QIV:

x₂ = 360^o 60^o = 300^o

Since we are supposed to express the solutions in terms of radians we get

x₁ = 2/3

and x₂ = 5/3

Example 7:

Solve sin x + 1 = 0 for x in the interval .  Express the solutions in degrees.

Step 1:

The interval in which we are supposed to find the solutions is. We will isolate the trigonometric ratio by adding 1 to both sides to get

sin x = 1

Step 2:

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

Step 3:

is a quadrantal angle, which has no reference angle!

Step 4: 

Since we do not have a reference angle, we are going to show the following picture:

From the picture above, we find that sin x = 1 only at 270^o on interval .

Therefore, we have one solution, namely x = 270^o.

Please note that the angle found in Step 2 is a perfectly good solution to the given equation. However, it is not part of the solution set because it does not lie in the solution interval [0^o, 360^o). 

Example 8:

Solve sin x 1 = 0 for x in the interval .  Express the solutions in degrees.

Step 1:

The interval in which we are supposed to find the solutions is . We will isolate the trigonometric ratio by adding 1 to both sides to get

sin x = 1

Step 2:

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

Step 3:

is a quadrantal angle, which has no reference angle!

Step 4: 

Since we do not have a reference angle, we are going to show the following picture:

From the picture above, we find that sin x = 1 only at 90^o on interval .

Therefore, we have one solution, namely .

However, we must find the solutions in the interval . We will simple add another 360^o rotation to 90^o to find a second solution, namely, .

Example 9:

Find ALL solutions for .  Express the solutions in EXACT radians.

Step 1:

We will find the solutions in the interval first. We will isolate the trigonometric ratio by dividing both sides by 2 to get

Step 2:

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

Helpful Hint: It's easier to work in degrees and then change the solutions back to radians! In this case, we will change the solution interval to degrees, namely .

Step 3:

Given the solution in Step 2, we will find its reference angle. Since 150^o is a QII angle, its reference angle is 180^o 150^o = 30^o.

Step 4:

The value of the cosine ratio in Step 1 is negative and we have a reference angle. Next, we will use All Students Take Calculus to find the quadrants in which the cosine ratio is negative. It is best to draw a picture!

We need to find solution angles located in QII and QIII with a reference angle of 30^o. We find the solutions for angle x in the interval to be as follows:

Solution in QII:

x₁ = 180^o 30^o = 150^o

Solution in QIII:

x₂ = 180^o + 30^o = 210^o

Since we are supposed to express the solutions in terms of radians we get

x₁ = 5/6

and x₂ = 7/6

We are asked to find ALL solutions expressed in radians. We see in Step 4 that the solutions are more than (or 180^o) apart,

Therefore, we add 2k to these solutions, where k is any integer.

That is, x₁ = 5/6 + 2k and x₂ = 7/6 + 2k.

Example 10:

Find ALL solutions for 2cos x = 0.  Express the solutions in degrees.

Step 1:

We will find the solutions in the interval first. We will isolate the trigonometric ratio by dividing both sides by 2 to get

cos x = 0

Step 2:

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

Step 3:

is a quadrantal angle, which has no reference angle!

Step 4: 

Since we do not have a reference angle, we are going to show the following picture:

From the picture above, we find that cos x = 0 at 90^o and 270^oon interval .

Therefore, we have two solutions, namely x₁ = 90^o and x₂ = 270^o.

We are asked to find ALL solutions expressed in degrees. We see in Step 4 that the cosine ratio is 0 at intervals of 180^o.

Therefore, to show all solutions, we only need to add 180^ok to the smaller of the two solutions in the interval between 0^o and 360^o where k is defined as any integer. 

That is, all solutions for angle x are 90^o + 180^ok.

Example 11:

Find ALL solutions for 3tan x = 0.  Express the solutions in degrees.

Step 1:

We will find the solutions in the interval first. We will isolate the trigonometric ratio by dividing both sides by 3 to get

tan x = 0

Step 2:

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

tan x = 0

x = tan^-1 0

x = 0^o

Step 3:

0^o is a quadrantal angle, which has no reference angle!

Step 4: 

Since we do not have a reference angle, we are going to show the following picture:

From the picture above, we find that tan x = 0 at 0^o and 180^o on interval . Please note that the solution of 360^o is excluded from the solution interval!

Therefore, we have two solutions, namely x₁ = 0^o and x₂ = 180^o.

We are asked to find ALL solutions expressed in degrees. We see in Step 4 that the tangent ratio is 0 at intervals of 180^o.

Therefore, to show all solutions, we only need to add 180^ok to the smaller of the two solutions in the interval between 0^o and 360^o where k is defined as any integer. 

That is, x = 0^o + 180^ok.