Solving Trigonometric Equations - Trigonometry

By the double angle identity, we can find

So to get the zeros, solve:

This means that any number that when doubled equals a multiple of 180 degrees is a zero. In this case that includes

But the question asks for the smallest positive roots which excludes the negative and zero roots, leaving 90, 180, 270

Given the multiple choice nature of the problem, the easiest way to solve would be to simply plug in each answer and find the one that does not work.

However, we want to learn the math within the problem. We begin solving the equation by factoring

We then divide.

We then must remember that our left side is equivalent to something simpler.

We can therefore substitute.

We then must consider the angles whose cosine is . The two angles within the first revolution of the unit circle are and , but since our angle is , we need to consider the second revolution, which also gives us and .

But since is equal to each of these angles, we must divide them by 2 to find our answers. Therefore, we have

Therefore, there is only one answer choice that does not belong.

We begin by getting the right side of the equation to equal zero.

Next we factor.

We then set each factor equal to zero and solve.

or

We then determine the angles that satisfy each solution within one revolution.

The angles and satisfy the first, and satisfies the second. Only is among our answer choices.

; Divide both sides by 2 to get

; take the inverse sine on both sides

; the left side reduces to x, so

At this point, either use a unit circle diagram or a calculator to find the value.

Keep in mind that the problem asks for all solutions between and .

If you use a calculator, you will only get as an answer.

So we need to find another angle that satisfies the equation .

; First use the double angle identity for .

; divide both sides by 2

; subtract the from both sides

; factor out the

; Now we have the product of two expressions is 0. This can only happen if one (or both) expressions are equal to 0. So let each expression equal 0.

or ;

or ; Take the inverse of each function for each expression.

or ; The second equation is not possible so gives no solution, but the first equation gives us:

The fastest way to solve this problem is to substitute a new variable. Let .

The equation now becomes:

So at what angles are the sine and cosine functions equal. This occurs at

You may be wondering, "Why did you include

if they're not between and ?"

The reason is because once we substitute back the original variable, we will have to divide by 2. This dividing by 2 will bring the last two answers within our range.

Dividing each answer by 2 gives us

We begin by substituting a new variable .

; Use the double angle identity for .

; subtract the from both sides.

; This expression can be factored.

; set each expression equal to 0.

or ; solve each equation for

or ; Since we sustituted a new variable we can see that if , then we must have . Since , that means .

This is important information because it tells us that when we solve both equations for u, our answers can go all the way up to not just .

So we get

Divide everything by 2 to get our final solutions

; We start by substituting a new variable. Let .

; Use the double angle identity for cosine

; the 1's cancel, so add to both sides

; factor out a from both terms.

; set each expression equal to 0.

or ; solve the second equation for sin u.

or ; take the inverse sine to solve for u (use a unit circle diagram or a calculator)

; multiply everything by 2 to solve for x.

; Notice that the last two solutions are not within our range . So the only solution is .

; First divide both sides of the equation by 4

; Next take the square root on both sides. Be careful. Remember that when YOU take a square root to solve an equation, the answer could be positive or negative. (If the square root was already a part of the equation, it usually only requires the positive square root. For example, the solutions to are 2 and -2, but if we plug in 4 into the function the answer is only 2.) So,

; we can separate this into two equations

and ; we get

and

; Divide both sides by 3

; Take the square root on both sides. Just as the previous question, when you take a square root the answer could be positive or negative.

; This can be written as two separate equations

and ; Take the inverse tangent

and

; The expression is similar to a quadratic expression and can be factored.

; set both expressions equal to 0. Since they are the same, the solutions will repeat, so I will only write it once.

; take the inverse tangent on both sides

; use the double angle identity for cosine

; distribute the 3 on the right side

; add the to both sides

; divide both sides by 8

; take the square root on both sides (Remember: it could be positive or negative)

; separate into two equations and take the inverse sine

and ; Use a calculator

and (calculator will give -0.659, but that is not in our range, so add to get 5.624)

The last two solutions are found using a unit circle. Since our x can be negative or positive this means that there is a corresponding value in all of the quadrants.

Similarly, we can get our last answer x= 3.801 as this is the x value in the third quadrant and is found by adding .

You are asked to find the first TWO times the the weight is 3 inches below its equilibrium point. Since it is below the equilibrium, that 3 will be negative. The y variable is used to represent position so we need to have .

; now we solve. First, divide by -6.

; At this point, it may help to substitute a new variable. Once you get more practice with this type of problem, you can skip the substitution.

Let , then

; using a calculator or a unit circle diagram, find the first TWO angles that give us a cosing value of 1/2.

; bring back the variable t

; divide each solution by 3

.

Similar to the previous question, we set up the equation with

; divide both sides by 3

; Use a substitution to make things easier

; take the inverse sine. Use a calculator or unit circle diagram to find the first THREE solutions (that is what the problem asked for).

; bring back the expression for u and separate your answers.

; solve each equation.

Notice that you will have to add for each equation, but all the other fractions have a denominator of 6.

; add to both sides of each equation.

; Divide by 4 in each equation

; And finally simplify

We can factor the original expression as . This gives us 2 factors that we need to individually set equal to 0.

The simpler one is:

. The two values for x that give a cosine of 0 are and .

Now set the second factor equal to 0:

add 1 to both sides

divide by 2

. The two values for x that give a cosine of are and .

In total we have 4 answers,

We can think of this equation as:

Then we can factor it as:

. Now we set both factors equal to 0 and solve.

The simpler one is . Sine is 0 when x is 0 or . would also work, although it is just another way of saying 0.

Now we will set the second factor equal to zero:

add to both sides

multiply both sides by

This must have no solution, because sine and cosine are never any larger than 1, and they are never 1 simultaneously. This means that our only answers are 0 and .

This expression can be factored as:

. We can solve by setting each factor equal to 0.

gives the answers since sine is 0 for those values of x.

To continue solving, set the second factor equal to 0:

add to both sides.

divide both sides by 2.

The values of x that give a sine of are and , so this equation has 4 solutions.

To solve, first add 3 to both sides:

square both sides

divide by 2

Consulting the unit circle, sine is positive when

There are multiple solution paths, including using the quadratic formula. This ends up looking a little complicated, and realizing that is necessary. Here is a different way to solve this.

First, re-write the expression by simplifying the second term, and subtracting from both sides

We can see that two of these expressions involve the square root of 3. We can factor by grouping if we change the order the expressions are written in:

We can factor from the first two expressions and from the second two:

Since both groups have a factor of , we can re-write this as:

This gives us two solutions:

and

If the tangent is 1, that means that sine and cosine have the same measure and sign, so

If the tangent is , that means that cosine and sine have opposite signs, sine is and cosine is . So,