**
SOLVING COMPLEX TRIGONOMETRIC EQUATIONS **

Example 1:

Solve

for2sin^{2}x 3sin x + 1 = 0on the interval . Express the solutions in degrees.x

Step 1:This equation is "quadratic" in form. That is, one variable (sin x) is raised to the second power and the other, same variable, is raised to the first power.

Nothing needs to be done because the equation only contains sine ratios.

Step 2:We are going to attempt factoring. To make factoring easier, we will make a little substitution. That is, we let

equal some letter, saysin x.kThen we rewrite

2sin^{2}x 3sin x + 1asIt can be factored as2k^{2}3k + 1..(2k1)(k1)Then it should be easy to see that we can write

2sin^{2}x 3sin x + 1 =(2sin x1)(sin x1) = 0Using the

Zero Product Principle, we will now set both factors equal to 0 and then solve for.x(1)

then2sin x1 = 0sin x =(2)

thensin x1 = 0sin x = 1We now have two simple trigonometric equations. We must try to find solutions for both. The solutions of

cos2x + 3sin x2 = 0will be a combination of both.

Steps 3 through 6:(1) Let's use the calculator to solve

. See Lesson 12 - Solving Simple Trigonometric Equations.sin x =Using the arcsine, we find

, which indicates ax = 30^{o}reference angleof.30^{o}Next, we will use

All Students Take Calculusto find the quadrants in which the sine ratio is positive. It is best to draw a picture!We need to find angles located in QI and QII that have a

reference angle of. We find the solutions for angle30^{o}on the interval to be as follows:xSolution in QI:

=x_{1}30^{o}Solution in QII:

=x_{2}180^{o}=30^{o}150^{o}(2) Let's use the calculator to solve

. See Lesson 12 - Solving Simple Trigonometric Equations.sin x = 1With the calculator in degrees and using the arcsine, we find

, which indicates ax = 90^{o}Quadrantal Angle, therefore, there is noreference angle.Next, we use the following picture:

We find one solution for angle

on the interval to bex.90^{o}In summary, the solutions of

on the interval are2sin^{2}x 3sin x + 1 = 0,30^{o}90, and^{o}.150^{o}

Example 2:

Example 3:Solve

for2cos^{2}x + 3sin x = 0on the interval . Express the solutions in degrees.x

Step 1:This equation is "quadratic" in form. However, it contains sine and cosine ratios. We must change the equation so that it contains either all cosines or all sines.

Remembering the

Pythagorean Identity, we can change the given equation to the following:cos^{2}x = 1 sin^{2}x

2(1 sin^{2}x) + 3sin x = 0Using the

Distributive Property:

2 2sin^{2}x + 3sin x = 0Writing in descending powers of

:sin x

2sin^{2}x + 3sin x + 2 = 0Finally, we will multiply both sides by

to get1.2sin^{2}x 3sin x 2 = 0

Step 2:We are going to attempt factoring. To make factoring easier, we will now make a little substitution. That is, we let

equal some letter, saysin x.kThen we rewrite

2sin^{2}x 3sin x 2asIt can be factored as2k.^{2}3k 2.(2k + 1)(k 2)Then it should be easy to see that we can write

2sin^{2}x 3sin x 2 =(2sin x + 1)(sin x2) = 0Using the

Zero Product Principle, we will now set both factors equal to 0 and then solve for.x(1)

then2sin x + 1 = 0sin x =(2)

thensin x2 = 0sin x = 2We now have two simple trigonometric equations. We must try to find solutions for both. The solutions of

will be a combination of both.2cos^{2}x + 3sin x = 0

Steps 3 through 6:(1) Let's use the calculator to solve

. See Lesson 12 - Solving Simple Trigonometric Equations.sin x =With the calculator in degrees and using the arcsine, we find

, which indicates ax =30^{o}reference angleof.30^{o}Next, we will use

All Students Take Calculusto find the quadrants in which the sine ratio is negative. It is best to draw a picture!We need to find angles located in QIII and QIV that have a

reference angle of. We find the solutions for the angle30^{o}on the interval to be as follows:xSolution in QIII:

=x_{1}+180^{o}=30^{o}210^{o}Solution in QIV:

=x_{2}360^{o}=30^{o}330^{o}(2) Let's use the calculator to solve

. See Lesson 12 - Solving Simple Trigonometric Equations.sin x = 2Using the arcsine, we get a

Domain Error!

Why?Please note that

is never greater thansin x(remember its graph?), therefore,1is undefined, therefore, has no solution..sin x = 2In summary, the solutions of

on the interval are2cos^{2}x + 3sin x = 0and210^{o}.330^{o}

Solve

sin^{2}x2cos xfor2 = 0on the interval . Express the solutions in degrees.x

Step 1:This equation is "quadratic" in form. However, it contains sine and cosine ratios. We must change the equation so that it contains either all cosines or all sines.

Remembering the

Pythagorean Identity, we can change the given equation to the following:sin^{2}x = 1 cos^{2}x

(1 cos^{2}x)2cos x2 = 0Writing in descending powers of cos x:

cos^{2}x2cos x1 = 0Finally, we will multiply both sides by

to get1.cos^{2}x + 2cos x + 1 = 0

Step 2:We are going to attempt factoring. To make factoring easier, we will make a little substitution. That is, we let

equal some letter, saycos x.kThen we rewrite

ascos^{2}x + 2cos x + 1. It can be factored ask^{2}+ 2k + 1.(k + 1)(k + 1)Then it should be easy to see that we can write

cos^{2}x + 2cos x + 1 =(cos x + 1)(cos x + 1) = 0We have two identical factors. We will use the

Zero Product Principleto set one equal to 0 and then solve for.x

thencos x + 1 = 0cos x =1We now have one simple trigonometric equation. We must try to find its solutions. They will be the solutions of

sin^{2}x2cos xas well.2 = 0

Steps 3 through 6:Let's use the calculator to solve

. See Lesson 12 - Solving Simple Trigonometric Equations.cos x =1With the calculator in degrees and using the arccosine, we find

, which indicates ax = 180^{o}Quadrantal Angle, therefore, there is noreference angle.Next, we use the following picture:

In summary, there is only one solution of

sin^{2}x2cos xon the interval , namely2 = 0=x.180^{o}

Example 4:

Solve

forcot x = 4 sin^{2}x cot xon the interval . Express the solutions in radians.x

Step 1:This equation contains sine and cotangent ratios with one of the cotangent ratios being connected to the squared sine ratio via multiplication. Therefore, we will NOT attempt to change the equation so that it contains either all sines or all cotangents.

Step 2:We are going to solve this equation by factoring out the greatest common factor which seems to be

.cot x

cot x = 4 sin^{2}x cot x

0 = 4 sin^{2}x cot xcot x

0 = cot x (4 sin^{2}x1)Using the

Zero Product Principle, we will now set both factors equal to 0 and then solve for.x(1)

cot x = 0(2)

4 sin^{2}x= 01then

andWe now have three simple trigonometric equations. We must try to find solutions for all of them. The solutions of

cot x = 4 sin^{2}x cot xwill be a combination of them all.

Steps 3 through 6:(1) Finding the angle

xforwith the calculator is cumbersome.cot x = 0Hopefully, we remember that the cotangent ratio is only 0 when the angle is quadrantal. We also hopefully remember that .

The sine ratio cannot equal 0 otherwise the cotangent ratio is undefined. Let's look at the following picture.

At which angles is the sine ratio NOT equal to 0? We find two solution for angle

on the interval , namelyxand/2.3/2(2) Let's use the calculator to solve

andsin x =. See Lesson 12 - Solving Simple Trigonometric Equations.sin x =With the calculator in degrees and using the arcsine, we find

andx = 30^{o}, which indicates ax =30^{o}reference angleofin both cases.30^{o}Next, we will use

All Students Take Calculusto find the quadrants in which the sine ratio is both positive and negative. It is best to draw a picture!and

We need to find angles located in QI, QII, QIII, and QIV that have a

reference angle of. We find the solutions for angle30^{o}on the interval to be as follows:xSolution in QI:

=x_{1}30^{o}/6Solution in QII:

=x_{2}180^{o }=30^{o}150^{o}5/6Solution in QIII:

=x_{2}180^{o}+=30^{o}210^{o}7/6Solution in QIV:

=x_{2}360^{o }=30^{o}330^{o}11/6In summary, the solutions of

cot x = 4 sin^{2}x cot xon the interval are

,/2,3/2,/6,5/6, and7/6.11/6

Example 5:

Solve for

on the interval . Express the solutions in degrees.x

Step 1:This equation is "quadratic" in form. However, it only contains a squared cosine ratio.

Step 2:We are going to solve this equation for

by using thecos xSquare Root Property.First, we must change the leading coefficient of the square trigonometric term to 1 by dividing both sides of the equation by 2.

We can now apply the

Square Root Propertyto . Subsequently we getWe now have two simple trigonometric equations, namely

and

We must try to find solutions for both. The solutions of will be a combination of both.

Steps 3 through 6:Let's use the calculator to solve and . See Lesson 12 - Solving Simple Trigonometric Equations.

With the calculator in degrees and using the arccosine, we find

andx = 45^{o}, which indicates ax = 135^{o}reference angleofin both cases.45^{o }Next, we will use

All Students Take Calculusto find the quadrants in which the cosine ratio is both positive and negative. It is best to draw a picture!and

We need to find angles located in QI, QII, QIII, and QIV that have a reference angle of

. We find the solutions for the angle45^{o}on the interval to be as follows:xSolution in QI:

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

=x_{2}=180^{o }45^{o }135^{o}Solution in QIII:

=x_{3}=180^{o }+ 45^{o }225^{o}Solution in QIV:

=x_{4}360^{o }=45^{o}315^{o}

In summary, the solutions of

on the interval are,45^{o}135,^{o}225, and^{o}.315^{o}

Example 6: (Trigonometric equations involving multiples of an angle)

Solve for

on the interval . Express the solutions in degrees.xHere we we have an equation were

= which is the same asbx.xPlease note that we will work with the angle through Step 5. Only in Step 6 will we find the solution(s) for angle

.x

Step 1:The solution interval for angle

is .xThe solution interval for angle is half of this, namely .

Steps 2 through 5:See Lesson 12 - Solving Simple Trigonometric Equations.

(2) Isolate the trigonometric expression on one side:

(3) We are going to use the concept of

Inverse Trigonometric Functionsto solve for the angleusing the calculator in degree mode.

Please note that you cannot at this point multiply both sides of the equation by 2 to get a solution for angle x!!!(4) The

reference anglefor angle equals45.^{o}(5) The value of the cosine ratio in (2) is positive and we have a

reference angle.We will use

All Students Take Calculusto find the quadrants in which the cosine ratio is positive. It is best to draw a picture!Finally, we are going to use the reference angle from Step 3. We need to find angles located in QI and QIV that have a reference angle of

.45^{o}Therefore, the solutions for angle on the interval are as follows:

=45^{o}and

=360^{o}=45^{o}315^{o}

Step 6:Since we are asked to solve for angle

, we must now multiply both sides of the above equations byxto get2In summary, the solutions of on the interval are

90and^{o}.630^{o}

Example 7: (Trigonometric equations involving multiples of an angle)

Solve for

on the interval . Express the solutions in EXACT radians.xHere we we have an equation were

=bx4x.Please note that we will work with the angle

4xthrough Step 5. Only in Step 6 will we find the solution(s) for angle.x

Step 1:The solution interval for angle

is .xThe solution interval for angle

is four times as large as this, namely .4x

Steps 2 through 5:See Lesson 12 - Solving Simple Trigonometric Equations.

(2) Isolate the trigonometric expression on one side:

(3) We are going to use the concept of

Inverse Trigonometric Functionsto solve for the angleusing the calculator.4xLet's work with degrees! Using the calculator in degree mode, we find

Please note that you cannot at this point divide both sides of the equation by 4 !!!(4) The

reference angleof angleequals4x30.^{o}(5) The value of the sine ratio in (2) is positive and we have a

reference angle.Qe will use

All Students Take Calculusto find the quadrants in which the sine ratio is positive. It is best to draw a picture!Finally, we are going to use the reference angle from Step 3. Now, remember the solution interval for angle

is in degrees.4xHowever, we are going to find angles in the solution interval first. They must be in QI and QII and have a reference angle of

.30^{o}They are

=4x_{1}30^{o}

=4x_{2}180^{o }=30^{o}150^{o}Since we are supposed to find solutions for angle

on the interval , we must add the following angles, which are coterminal to the two we already found:4x

=4x_{3}+30^{o}=360^{o}390^{o}

=4x_{4}150^{o }+=360^{o}510^{o}Let's go ahead and express these four solutions in radians as follows:

Step 6:Since we are asked to solve for the angle

, we must now divide both sides of the above equations byxto get4

In summary, the solutions of on the interval are

.