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 45o is a QI angle, its reference angle equals 45o.

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 45o. We find the solutions for angle x in the interval to be as follows.

Solution in QI:

x1 = 45o

Solution in QII:

x2 = 180o 45o = 135o

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 45o is a QI angle, its reference angle equals 45o. Therefore, the reference angle of 45o is 45o 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 45o. We find the solutions for angle x in the interval to be as follows.

Solution in QIII:

x1 = 180o + 45o = 225o

Solution in QIV:

x2 = 360o 45o = 315o

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 60o is a QI angle, its reference angle equals 60o.

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 60o. We find the solutions for angle x in the interval to be as follows.

Solution in QI:

x1 = 60o

Solution in QIII:

x2 = 180o + 60o = 240o

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 [180o,180o).

Step 3:

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

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 60o.

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

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

Solution in QI:

x1 = 60o

Solution in QIII:

x2 = 180o + 60o = 120o

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 [265o, 90o] in degrees.  Round all calculations to 2 decimal places.

Step 1:

The interval in which we are supposed to find the solutions is [265o, 90o]. 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 149.09o

Step 3:

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

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 30.91o.

Here Step 4 it a bit more complicated because we are asked to find solutions in the interval [265o, 90o]. 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 [265o, 90o].

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

Solution in QII:

x1 = [ 30.91o + 180o] = 210.91o

Solution in QIII:

x2 = 180o + 30.91o = 149.09o

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 60o is a QI angle, its reference angle equals 60o. Therefore, the reference angle of 60o is 60o 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 60o. We find the solutions for angle x in the interval to be as follows:

Solution in QII:

x1 = 180o 60o = 120o

Solution in QIV:

x2 = 360o 60o = 300o

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 270o on interval .

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

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 [0o, 360o)

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 90o on interval .

Therefore, we have one solution, namely .

However, we must find the solutions in the interval . We will simple add another 360o rotation to 90o 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 150o is a QII angle, its reference angle is 180o 150o = 30o.

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 30o. We find the solutions for angle x in the interval to be as follows:

Solution in QII:

x1 = 180o 30o = 150o

Solution in QIII:

x2 = 180o + 30o = 210o

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 180o) 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 90o and 270oon interval .

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

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

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

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

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 = 0o

Step 3:

0o 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 0o and 180o on interval . Please note that the solution of 360o is excluded from the solution interval!

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

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

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

That is, x = 0o + 180ok.