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Example Questions
Example Question #1 : Finding Trigonometric Roots
Which of the following is a solution to the following equation such that
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.
Example Question #1 : Finding Trigonometric Roots
Solve the following equation for .
No solution exists
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
Example Question #11 : Solving Trigonometric Equations
Solve the equation for .
No solution exists
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
Example Question #1 : Finding Trigonometric Roots
Solve the following equation for .
; 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
.
Example Question #1 : Finding Trigonometric Roots
Solve the following equation for .
No solution exists
; 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
Example Question #1 : Finding Trigonometric Roots
Solve the equation for .
; 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
Example Question #1 : Finding Trigonometric Roots
Solve the following equation for .
; 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
Example Question #1 : Finding Trigonometric Roots
Solve the equation below for greater than or equal to
and strictly less than
.
and
only
and
only
and
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
.
Example Question #9 : Finding Trigonometric Roots
Solve the following equation. Find all solutions such that .
No Solution
; 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 .
Example Question #1 : Finding Trigonometric Roots
Solve the following equation. Find all solutions such that .
No solution exists
; 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:
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