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

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

.