Trigonometric Equations - Trigonometry

Therefore,

and that only happens once in the given interval, at , or 45 degrees.

The fastest way to solve this equation is to simply try the three answers. Plugging in gives

Our first choice is valid.

Plugging in gives

However, since is undefined, this cannot be a valid answer.

Finally, plugging in gives

Therefore, our third answer choice is not correct, meaning only 1 is correct.

Note first that:

and :

.

Now taking . We have

.

Since and .

We therefore have :

We have .

Now since

This last expression can be written as :

.

This shows the required result.

Letting , we have the equivalent expression:

.

We cant factor since .

This shows that we cannot factor the above expression.

We first note that we have:

Then taking , we have the result.

First of all we know that :

and this gives:

.

Now we need to see that: can be written as

and since

we have then:

.

We know that we can write

in the following form

.

Now taking ,

we have:

.

This is the result that we need.

From the expression :

we have:

Now since we know that :

. This expression becomes:

.

This is what we need to show.

First we see that :

.

Now letting

we have

We know that :

and we are given that

, this gives

Step 1: Recall the difference of squares (or powers of four) formula:

Step 2: Factor the question:

Factor more:

Step 3: Recall a trigonometric identity:

.. Replace this

Final Answer:

Recall the values of for which . If it helps, think of sine as the values on the unit circle. Thus, the acceptable values of would be 0, 180, 360, 540 etc.. However, in our scenario .

Thus we have and .

Any other answer would give us values greater than 90. When we divide by 4, we get our answers,

and .

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