Example 2:

Find the general solution of for x.  Express it in EXACT radians.

Whenever we are asked to find generalsolutions, we first find the solutions in the interval . We already did this in Example 1.

We found x₁ = /3 (60^o) and x₂ = 4/3 (240^o).

We note that these two solutions are exactly (180^o) apart. Therefore, we will only add k the smallest solution.

That is, the general solution of is /3 + k.

Example 3:

We are going to solve again, but now we only want solutions for x in the interval .  Express the solutions in EXACT radians. Compare with Example 1!

Step 1:

The solution interval for angle x is . The trigonometric ratio is already isolated on one side of the equal sign. 

Therefore, we use .

Step 2:

Given , we find x = tan ^-1 ()

Helpful Hint: It is often easier to work in degrees and then change the solutions back to radians. So lrt's change the solution interval to degrees, namely , and then we continue to work with degrees.

With the calculator in degree mode, we get x = 60^o

Step 3:

Given a solution for angle x 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:

We have a reference angle and the value of the tangent ratio in Step 1 is positive.

Therefore, we will use All Students Take Calculus to find the quadrants in which the tangent ratio is positive. See picture below!

 

Finally, let's first indicat the solution intervale for angle x by graphing the angles 180^o and 180^o. Then we draw a reference angle into the appropriate quadrants.

We now find the solutions for angle x in the interval 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 4:

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

Step 1:

The solution interval for angle x is .  We will then isolate the trigonometric ratio by dividing both sides of the equal sign by 1.

We end up with .

Step 2:

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

Given , we find we find x = tan ^-1 ()

Helpful Hint: It is often easier to work in degrees and then change the solutions back to radians. So let's change the solution interval to degrees, namely , and then we continue to work with degrees.

With the calculator in degree mode, we get x = 60^o.

Step 3:

Given a solution for angle x 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:

We have a reference angle and the value of the tangent ratio in Step 1 is negative.

Therefore, we will use All Students Take Calculus to find the quadrants in which the tangent ratio is positive. Thus, the tangent is negative in QII and QVI.  See picture below!

Finally, let’s first indicate the solution interval for angle x by graphing the angle 360^o.  Then we draw the reference angle in the appropriate quadrants.

We now find the solutions for angle x in the interval with the help of the picture above.

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 5:

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

Step 1:

The solution interval for angle x is .  We will then isolate the trigonometric ratio by dividing both sides of the equal sign by 2.

We end up with .

Step 2:

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

Given , we find .

Helpful Hint: It's easier to work in degrees and then change the solutions back to radians. So let's change the solution interval to degrees, namely , and then we continue to work with degrees.

With the calculator in degree mode, we get x = 45^o.

Step 3:

Given a solution for angle x 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:  

We have a reference angle and the value of the sine ratio in Step 1 is positive.

Therefore, we will use All Students Take Calculus to find the quadrants in which the sine ratio is positive. See picture below!

 

Finally, let’s first indicate the solution interval for angle x by graphing the angle 360^o.  Then we draw the reference angle in the appropriate quadrants.

We now find the solutions for angle x in the interval with the help of the picture above.

Solution in QI:

x₁ = 45^o

Solution in QII:

x₂ = 180^o45^o = 135^o

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 solution interval for angle x is .  We will then isolate the trigonometric ratio by dividing both sides of the equal sign by 2.

We end up with .

Step 2:

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

Given , we find .

Helpful Hint: It's easier to work in degrees and then change the solutions back to radians. So, let’s change the solution interval to degrees, namely , and then we continue to work with degrees.

With the calculator in degree mode, we get x = 45^o.

Step 3:

Given a solution for angle x 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:

We have a reference angle and the value of the sine ratio in Step 1 is negative.

Therefore, we will use All Students Take Calculus to find the quadrants in which the sine ratio is positive. Thus, the sine is negative in QIII and QVI.  See picture below!

Finally, let’s first indicate the solution interval for angle x by graphing the angle 360^o.  Then we draw the reference angle in the appropriate quadrants.

We now find the solutions for angle x in the interval with the help of the picture above.

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 7:

Solve cos x = 0.858 for x in the interval in radians.  Round all calculations to 3 decimal places.

Step 1:

The solution interval for angle x is . The trigonometric ratio is already isolated on one side of the equal sign. 

We use 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. 

Note: Here it is not a good idea to work in degrees. The final solutions would become more and more inexact since we need to round to find the angle and then do more rounding to change from degrees to radians.

Given cos x = 0.858, we find x = cos^-1 (0.858).

With the calculator in radian mode, we get x 2.602.

Step 3:

Given a solution for angle x in Step 2, we will find its reference angle. Since 2.602 (149^o) is a QII angle, its reference angle is 2.602 0.540.

Step 4:

We have a reference angle and the value of the cosine ratio in Step 1 is negative.

Therefore, we will use All Students Take Calculus to find the quadrants in which the cosine ratio is positive. Thus, the cosine is negative in QII and QIII.  See picture below!

Finally, let’s first indicate the solution interval for angle x by graphing the angle .  Then we draw the reference angle in the appropriate quadrants.

We can now find the solutions for angle x in the interval with the help of the picture above.

Solution in QII:

x₁ 2.602

Solution in QIII:

x₂ + 0.540 3.646

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 of the equal sign.

We end up with 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. 

Given , we get x = sin ^{- 1}(1)

With the calculator in degee mode, we find x = 90^o.

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 does not lie in the solution interval [0^o, 360^o). Therefore, it is not a solution in our case.

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 of the equal sign.

We end up with 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. 

Given , we get we get x = sin ^-1(1)

Therefore, with the calculator in radian mode, we find x = 90^o.

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 10:

Find the general solution for .  Express it in EXACT radians.

Step 1:

We will find the solutions in the interval first. We will isolate the trigonometric ratio by dividing both sidesof the equal sign by 2.

We end up with .

Step 2:

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

Given , we get .

Helpful Hint: It's easier to work in degrees and then change the solutions back to radians. So, let’s change the solution interval to degrees, namely , and then we continue to work with degrees.

With the calculator in degree mode, we get x = 150^o.

Step 3:

Given a solution for angle x 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:

We have a reference angle and the value of the cosine ratio in Step 1 is negative.

Therefore, we will use All Students Take Calculus to find the quadrants in which the cosine ratio is positive. Thus, the cosine is negative in QII and QIII.  See picture below!

Finally, let’s first indicate the solution interval for angle x by graphing the angle 360^o.  Then we draw the reference angle in the appropriate quadrants.

We now find the solutions for angle x in the interval with the help of the picture above.

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 the general solution expressed in radians. We see in Step 4 that the solutions are NOT exactly (or 180^o) apart.

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

The general solution for angle x is as follows:

5/6 + 2k and 7/6 + 2k

Example 11:

Find the general solution for 2cos x = 0.  Express it in degrees.

Step 1:

We will find the solutions in the interval first. We will isolate the trigonometric ratio by dividing both sides of the equal sign by 2.

We end up with 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. 

Given , we get x = cod ^-1(0).

Therefore, with the calculator in degree mode, we find x = 90^o.

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 the general solutions expressed in degrees. We note that these two solutions are exactly 180^o apart.

Therefore, to show the general 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. 

The general solution for angle x is 90^o + 180^ok.

Example 12:

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

Step 1:

We will find the solutions in the interval first. We will isolate the trigonometric ratio by dividing both sides of the equal sign by 3.

We end up with 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. 

Given tan x = 0, we get x = tan^-1 (0)

Therefore, with the calculator in degree mode, we find 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 because the solution interval excludes this angle since there is a parenthesis next to it 360^o !!!

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

We are asked to find the general solutions expressed in degrees. We note that these two solutions are exactly 180^o apart.

Therefore, to show the general solution, 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. 

The general solution for angle x is 0^o + 180^ok.