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 . The trigonometric ratio is already isolated on one side of the equal sign.
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!
Let's draw the reference angle in these quadrants as well as the solution interval .
We now find the solutions for angle x in the interval with the help of the picture above.
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 2:
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.
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!
Let's draw the reference angle in these quadrants as well as the solution interval.
We now find the solutions for angle x in the interval 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 3:
Find ALL solutions of for x. Express the solutions in EXACT radians.
Whenever we are asked to find ALL solutions, we first find the solutions in the interval . We already did this in Example 1.
We found x1 = /3 and x2 = 4/3.
Please note that these two solutions are exactly apart. Therefore, we will only add k the smallest solution.
That is, ALL solutions of are /3 + k.
Example 4:
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!
Let's draw the reference angle in these quadrants as well as the solution interval .
We now find the solutions for angle x in the interval with the help of the picture above.
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 5:
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
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!
Let's draw the reference angle in these quadrants as well as the solution interval .
We now find the solutions for angle x in the interval with the help of the picture above.
Solution in QI:
x1 = 45o
Solution in QII:
x2 = 180o45o = 135o
Since we are supposed to express the solutions in terms of radians we get
x1 = /4
and x2 = 3/4
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 now 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. 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!
Let's draw the reference angle in these quadrants as well as the solution interval .
We now find the solutions for angle x in the interval with the help of the picture above.
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 7:
Solve cos x = 0.858 for x in the interval in degrees. Round all calculations to 2 decimal places.
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.
cos x = 0.858
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!
Let's draw the reference angle in these quadrants as well as the solution interval .
We now find the solutions for angle x in the interval 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 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 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 9:
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 = 1Step 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 90oto find a second solution, namely,.
Example 10:
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!
Let's draw the reference angle in these quadrants as well as the solution interval .
We now find the solutions for angle x in the interval with the help of the picture above.
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 11:
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 12:
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 = 0Step 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.