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Example Questions
Example Question #221 : Trigonometry
Find the roots of the equation
Factor the equation:
The roots occur when each factor equals 0:
Example Question #222 : Trigonometry
Find all roots of this equation, in radians.
Take the square root of both sides:
This gives us two roots,
and
Solving for x:
or ,
which evaluate to and
Now we check to make sure both answers work. Make sure your calculator is in radians mode!
Our answers are
Example Question #223 : Trigonometry
If What is in degrees?
if is
Example Question #1 : Quadratic Formula With Trigonometry
Solve the following equation for .
No solution exists
; Use the double angle identity for cosine.
; Move everything to the left side of the equation.
; This is a quadratic-like expression that cannot be factored. We must use the quadratic formula. It may be helpful to see this if you replace with , so it becomes:
Recall the quadratic formula
plug in .
We now have
; Separate this into two equations and take the inverse sine.
or
The first equation gives us . Using the unit circle as we did in previous problems, we can find a second answer from this which is . The second equation will not give us a solution.
Example Question #2 : Quadratic Formula With Trigonometry
What are the zeros of the function listed above for the interval .
When the quadratic formula is applied to the function, it yields
So those are the zeros for sine, but sine has a minimum of -1, so -2 is out. For -1/2, sine achieves that twice in a cycle, at π+π/6 and 2π-π/6. So while -π/6 is true, it is not correct since it is not in the given interval.
Therefore on the given interval the zeros are:
Example Question #3 : Quadratic Formula With Trigonometry
Solve the following trigonometric equation:
for
The equation does not have a solution.
Since can be written as:
. We can't have .
Therefore . This means that where k is an integer.
since . We have x=0 is the only number that satisfies this property.
Example Question #1 : Quadratic Formula With Trigonometry
Solve each equation over the domain (answer in degrees).
Rearrange the problem,
Over the interval 0 to 360 degrees, cosx = 1/2 at 60 degrees and 300 degrees.
Example Question #2 : Quadratic Formula With Trigonometry
Solve the equation over the interval
First, get the equation in terms of one trig function. We can do this by substituting in using the Pythagorean Identity for .
Then we have .
Bring all the terms to one side to find that .
We can factor this quadratic to .
This means that .
The only angle value for which this is true is .
Example Question #2 : Quadratic Formula With Trigonometry
Solve for , giving your answer as a positive angle measure:
No solution
First, re-write the equation so that it is equal to zero:
Now we can use the quadratic formula to solve for x. In this case, the coefficients a, b, and c are a=1, b=2, and c=-3:
simplify
This gives two potential answers:
and
Sine must be between -1 and 1, so there are no values of x that would give a sine of -3. The only solution that works is . The only angle measure that has a sine of 1 is .
Example Question #2 : Quadratic Formula With Trigonometry
Solve for :
.
Give your answer as a positive angle measure.
Use the quadratic formula to solve for x. In this case, the coefficients a, b, and c are a=4, b=1, and c=-1:
simplify
the square root of 17 is about 4.123. This gives two potential answers:
. We can solve for x by evaluating both and . The first gives an answer of . Add this to 360 to get that as a positive angle measure, . If this has a sine of -0.64, so does its reflection over the y-axis, which is .
The second gives an answer of . If that has a sine of 0.39, then so does its reflection over the y-axis, which is .
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