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SOLVING SIMPLE TRIGONOMETRIC EQUATIONS **

Example 1:

Solve for

in the interval . Express the solutions inxEXACTradians.

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

to get2

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

Step 2:We are going to use the concept of

Inverse Trigonometric Functionsto solve for the angleusing the calculator.x

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 angleis a QI angle, its45^{o}reference angleequals.45^{o}

Step 4:The value of the sine ratio in Step 1 is positive and we have a

reference angle. Next, we will useAll Students Take Calculusto find the quadrantsin 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 angleof. We find the solutions for angle45^{o}in the intervalxto be as follows.Solution in QI:

=x_{1}45^{o}Solution in QII:

=x_{2}180^{o}=45^{o}135^{o}Since we are supposed to express the solutions in terms of radians we get

=x_{1}/4and

=x_{2}3/4

Example 2:

Solve for

in the interval . Express the solutions inxEXACTradians.

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

to get2

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

Step 2:We are going to use the concept of

Inverse Trigonometric Functionsto solve for the angleusing the calculator.x

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 r

eference angle.We know that thereference angleof a negative angle is equal to that of its positive counterpart.Since

is a QI angle, its45^{o}reference angleequals. Therefore, the45^{o}reference angleofis45^{o}as well.45^{o}

Step 4:The value of the sine ratio in Step 1 is negative and we have a

reference angle. Next, we will useAll Students Take Calculusto find the quadrantsin 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 angleof. We find the solutions for angle45^{o}in the interval to be as follows.xSolution in QIII:

=x_{1}+180^{o}=45^{o}225^{o}Solution in QIV:

=x_{2}360^{o}=45^{o}315^{o}Since we are supposed to express the solutions in terms of radians we get

=x_{1}5/4and

=x_{2}7/4

Example 3:

Solve for

in the interval . Express the solutions inxEXACTradians.

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 Functionsto solve for the angleusing the calculator.x

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 r

eference angle.Since angleis a QI angle, its60^{o}reference angleequals.60^{o}

Step 4:The value of the tangent ratio in Step 1 is positive and we have a

reference angle. Next, we will useAll Students Take Calculusto find the quadrantsin 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 angleof. We find the solutions for angle60^{o}in the interval to be as follows.xSolution in QI:

=x_{1}60^{o}Solution in QIII:

=x_{2}180^{o}+=60^{o}240^{o}Since we are supposed to express the solutions in terms of radians we get

=x_{1}/3and

=x_{2}4/3

Example 4:

We are going to solve again, but now we only want solutions for

in the interval . Express the solutions inxEXACTradians.

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:Inverse Trigonometric Functionsto solve for the angleusing the calculator.x

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 r

eference angle.Since angleis a QI angle, its60^{o}reference angleequals.60^{o}

Step 4:The value of the tangent ratio in Step 1 is positive and we have a

reference angle. Next, we will useAll Students Take Calculusto find the quadrantsin 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 angleof.60^{o}

Here Step 4 is a bit more complicated because we are asked to find solutions in the interval[180. Drawing a picture is always a good idea.^{o},180^{o})Since we determined that the solutions are in QI and QIII, let's draw the

reference anglesin these quadrants AND the solution interval.[180^{o},180^{o})We now find the solutions for angle

in the intervalxwith the help of the picture above.[180,^{o}180^{o}]Solution in QI:

=x_{1}60^{o}Solution in QIII:

=x_{2}180^{o}+60^{o}=120^{o}Since we are supposed to express the solutions in terms of radians we get

=x_{1}/3and

=x_{2}2/3

Example 5:

Solve

forcos x = 0.858in the intervalx[,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

[26,5^{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:Inverse Trigonometric Functionsto solve for the angleusing the calculator.x

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 r

eference angle. Sinceis a QII angle, its149.09^{o}reference angleis.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 useAll Students Take Calculusto find the quadrantsin 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 angleof.30.91^{o}

Here Step 4 it a bit more complicated because we are asked to find solutions in the interval. Drawing a picture is always a good idea.[,265^{o}90^{o}]Since we determined that the solutions are in QII and QIII, let's draw the

reference anglesin these quadrants and the solution interval[,265^{o}90^{o}.]We now find the solutions for angle

in the intervalxwith the help of the picture above.[,265^{o}90^{o}]Solution in QII:

=x_{1}[+30.91^{o}180^{o}]=210.91^{o}Solution in QIII:

=x_{2}+180^{o}30.91^{o}=149.09^{o}

Example 6:

Solve for

in the interval . Express the solutions inxEXACTradians.

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

to get1

Step 2:Inverse Trigonometric Functionsto solve for the angleusing the calculator.x

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 r

eference angle.We know that thereference angleof a negative angle is equal to that of its positive counterpart.Since

is a QI angle, its60^{o}reference angleequals. Therefore, the60^{o}reference angleofis60^{o}as well.60^{o}

Step 4:The value of the tangent ratio in Step 1 is negative and we have a

reference angle. Next, we will useAll Students Take Calculusto find the quadrantsin 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 angleof. We find the solutions for angle60^{o}in the interval to be as follows:xSolution in QII:

=x_{1}180^{o}=60^{o}120^{o}Solution in QIV:

=x_{2}360^{o}=60^{o}300^{o}Since we are supposed to express the solutions in terms of radians we get

=x_{1}2/3and

=x_{2}5/3

Example 7:

Solve

forsin x + 1 = 0in the interval . Express the solutions in degrees.x

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

to both sides to get1

sin x=1

Step 2:

Inverse Trigonometric Functionsto solve for the angleusing the calculator.x

Step 3:

is aquadrantal angle, which has noreference 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=only at1on interval .270^{o}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 xfor1 = 0in the interval . Express the solutions in degrees.x

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

to both sides to get1sin x= 1

Step 2:Inverse Trigonometric Functionsto solve for the angleusing the calculator.x

Step 3:is a

quadrantal angle, which has noreference 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 xonly at= 1on interval .90^{o}Therefore, we have one solution, namely .

However, we must find the solutions in the interval . We will simple add another

rotation to360^{o}to find a second solution, namely, .90^{o }

Example 9:

Find ALL solutions for . Express the solutions in

EXACTradians.

Step 1:We will find the solutions in the interval first. We will isolate the trigonometric ratio by dividing both sides by

to get2

Step 2:Inverse Trigonometric Functionsto solve for the angleusing the calculator.x

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 r

eference angle. Sinceis a QII angle, its150^{o}reference angleis.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 useAll Students Take Calculusto find the quadrantsin 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 angleof. We find the solutions for angle30^{o}in the interval to be as follows:xSolution in QII:

=x_{1}180^{o}=30^{o}150^{o}Solution in QIII:

=x_{2}180^{o}=+30^{o}210^{o}Since we are supposed to express the solutions in terms of radians we get

=x_{1}5/6and

=x_{2}7/6

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

to these solutions, where2kis any integer.kThat is,

=x_{1}+5/6and2k=x_{2}7/6+.2k

Example 10:

Find ALL solutions for

. Express the solutions in degrees.2cos x = 0

Step 1:We will find the solutions in the interval first. We will isolate the trigonometric ratio by dividing both sides by

to get2

cos x= 0

Step 2:Inverse Trigonometric Functionsto solve for the angleusing the calculator.x

Step 3:is a

quadrantal angle, which has noreference 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 xat= 0and90^{o}on interval .270^{o}Therefore, we have two solutions, namely

=x_{1}9and0^{o}=x_{2}.270^{o}

We are asked to find ALL solutions expressed in degrees.We see in Step 4 that the cosine ratio isat intervals of0.180^{o}Therefore, to show all solutions, we only need to add

to the smaller of the two solutions in the interval between180^{o}kand0^{o}where360^{o}is defined as any integer.kThat is, all solutions for angle

arex+90^{o}180^{o}.k

Example 11:

Find ALL solutions for

. Express the solutions in degrees.3tan x = 0

Step 1:We will find the solutions in the interval first. We will isolate the trigonometric ratio by dividing both sides by

to get3tan x= 0

Step 2:Inverse Trigonometric Functionsto solve for the angleusing the calculator.x

tan x = 0

x = tan^{-1}0

x = 0^{o}

Step 3:

is a0^{o}quadrantal angle, which has noreference 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

attan x = 0and0^{o}on interval . Please note that the solution of180^{o}is excluded from the solution interval!360^{o}Therefore, we have two solutions, namely

=x_{1}and0^{o}=x_{2}.180^{o}

We are asked to find ALL solutions expressed in degrees.We see in Step 4 that the tangent ratio isat intervals of0.180^{o}Therefore, to show all solutions, we only need to add

to the smaller of the two solutions in the interval between180^{o}kand0^{o}where360^{o}is defined as any integer.kThat is,

=x+0^{o}180^{o}.k