All Trigonometry Resources
Example Questions
Example Question #23 : Solving Trigonometric Equations
The equation represents the motion of a weight hanging on a spring after it has been pulled 6 inches below its equilibrium point and released. The variable represents the weight's position above or below the equilibrium point. For example, means the weight is 5 inches below its equilibrium point. means the weight is 5 inches above its equilibrium point. The variable represents the time, in seconds, after the weight was released.
Find the first three times the weight is 1.5 inches above its equilibrium point.
None of the other answers
No solution exists
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
Example Question #211 : Trigonometry
Find the three smallest positive roots of the above equation.
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
Example Question #25 : Solving Trigonometric Equations
Which of the following is NOT a solution to the equation below such that ?
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.
Example Question #51 : Trigonometric Equations
Solve for . Give your answer in radians.
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,
Example Question #52 : Trigonometric Equations
Solve for . Give your answer in radians.
No solution
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 .
Example Question #53 : Trigonometric Equations
Solve for , giving an answer in radians.
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.
Example Question #54 : Trigonometric Equations
Solve for :
To solve, first add 3 to both sides:
square both sides
divide by 2
Consulting the unit circle, sine is positive when
Example Question #55 : Trigonometric Equations
Solve for :
No solution
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,
Example Question #219 : Trigonometry
Solve for using a trigonometric identity:
First, using the trigonometric identity , re-write the expression so it is entirely in terms of :
if we add to both sides and re-order the terms, we can see that this is a quadratic equation in terms of sine:
so we can use the quadratic formula to solve this way:
This gives us two potential answers:
this is outside of the range of sine, so it won't work.
Taking the inverse on a calculator yields , which we can add to 360 to give .
That's just one answer, however. To get the other answer, we need to know the other angle below the x-axis, so we can add.
Example Question #220 : Trigonometry
Find the roots of the equation
We can factor this expression by grouping.
re-arrange the original expression so that the terms with sine are next to each other
factor our the
this factors to
Since we are finding the roots, we should set both factors equal to 0:
add 1 to both sides
subtract 1 from both sides
divide by 2
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