;  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

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 . 

; 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:

;  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

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.