Download Solving Equations Tricks tutorial

Advanced Solving Techniques
We have already discussed solving simple linear and quadratic models of trigonometric functions.
*** Remember:::: set the equation = 0 to find out whether you want to use the
star method to factor or the GCF to factor
So… What happens when these two factoring methods above do not work????
We need a different set of tools.
Technique #1
The QUADRATIC FORMULA
What formula did you use when you could not factor a quadratic equation? You’re right!
---The Quadratic Formula---
 b  b 2  4ac
2a
This was great when you wanted to solve for x, but what if you need to solve for the cscx????
x
Example:
csc2x + 3cscx = 5
csc2x + 3cscsx – 5 = 0
set it = 0 first
because the star method fails here, you must modify the quadratic
formula to fit your needs
so… Let x = cscx. . .then… the equation becomes x2 + 3x – 5 = 0.
a = 1, b= 3, c= -5 , but x in the formula becomes cscx
csc x 
 3  32  4(1)( 5)
2(1)
plug values of a, b, and c into the formula
 3  29
2
cscx = 1.19 ,-4.19
simplify the values
So…. x = csc-1(1.19)
x = csc-1(-4.19)
solve for x
csc x 
use calculator to get number values
This means that the angles we are looking for are where the cscx is positive (QI, QII) and
the cscx is negative (QIII, QIV). You will need 4 answers to get this correct.
x = 57 + 360 n and 123 + 360 n
find the reference angle for QI and QII
Calculate sin-1(1/1.19) and use it to
x = 373.8 + 360 n and 193.8 + 360 n
find the reference angle for QIII and QIV
Calculate sin-1(1/-4.19) and use it to
The Quadratic Formula can be modified for any of the six trig functions. It is important
to note that this is useful ONLY when you have a single trig function in your equation.
Technique #2
IDENTITY SUBSTITUTION
How do you work with more than one function in an equation if factoring a GCF does not
work?
Example: 2 sin 2 x  3 cos x  3
2 sin 2 x  3 cos x  3  0
Set the equation = 0 first
To accommodate multiple functions, in this case sinx and cosx, we use an identity
substitution. The goal in substituting is to convert the equation to a single function rather
than work with multiple functions. BE CAREFUL though, sometimes factoring a GCF
will get the job done more efficiently.
So…. we will substitute 1 – cos2x for sin2x … now we have a single function.
2(1 – cos2x) + 3cosx – 3 = 0
Substitute the identity
2 – 2cos2x + 3cosx – 3 = 0
Distribute
-2cos2x + 3cosx – 1 = 0
Put in proper form
2
2cos x - 3cosx + 1 = 0
Multiply each term by -1
You now have a proper quadratic with a single function. If it factors, use the star method.
If not, you must resort to using the Quadratic Formula above.
Technique #3
SQUARING BOTH SIDES
Sometimes, when it looks like there is nothing that can be done, we can square both sides
of the equation.
Example:
cosx + 1 = sinx
Because this is not quadratic to begin with, the only way to get it to a quadratic is by
squaring. That is why this technique is useful.
(cosx + 1)2 = (sinx)2
cos2x + 2cosx + 1 = sin2x
cos2x + 2cosx + 1 – sin2x = 0
cos2x + 2cosx + 1 – (1 – cos2x) = 0
cos2x + 2cosx + 1 – 1 + cos2x = 0
2cos2x + 2cosx = 0
Square both sides
Foil and simplify
Set = 0
Substitute an identity
Distribute the negative
Combine like terms
You can now factor the GCF or use the star method to finish the problem.
Technique #4
MULTIPLE OR HALF ANGLES
Sometimes the function throws a twist at you by giving you a trig expression like
Sin 2x or cos (x/2)
or
tan 3x
These are called multiple angles.
We can easily use our previous notes to understand how to deal with these.
Example:
So…
2 cos(3t ) – 1 = 0
2 cos u – 1 = 0
2cos u = 1
cos u = ½
We simply Let u = 3t .
Substitue u for 3t
add 1 to both sides
divide both sides by 2
now… u = cos-1(1/2)
solve for u
since cos u is positive in QI and QIV, we understand that
u = 60 + 360 n
and
u = 300 + 360 n
***The trick to doing it this way, is that you must replace u with its orginal value at the
end and solve. Since we let u = 3t, we now have to replace u with 3t. ***
3t = 60 + 360 n
and
3t = 300 + 360 n
t = 20 + 120 n
and
t = 100 + 120 n
We now solve for t.
The key to doing this correctly is to substitute the multiple angle with a single variable and solve,
then replace the single variable at the end with its multiple angle and solve again.