2 types of Trigonometric Equations:
#1. Identities – equations that are valid for all values of the variable for which the
equation is defined. Ex. sin²x + cos²x = 1
ex. cot = 1/tanx
#2. Conditional Equations – valid only for certain values of the variable
Ex. sinx = 0
ex. cosx = ½
** solutions are found by—using methods similar to solving algebraic equations
8.1 Graphical Solutions to Trigonometric Equations
** use same methods as before except  trig. Equations typically have an infinite # of
solutions—determined by using the periodicity of
the function.
Basic Equation = an equation that involves a single trig. function set = to a #
(ex. sin x = .39,
cos x = .5)
Methods to solve trig. equations graphically:
#1. Intersection Method – given an equation put left side in y1 and right side in y2
Ex. tan x = -1
Steps: 1. put y1 = tan x and y2 = -1
Step 2: find the points of intersection – will see infinitely many
Step 3: Observe 1 complete cycle [-π/2. π/2]
** look at solution in that interval (change window to –π/2, to π/2)
X = -.80 tan repeats itself from left to right so other solutions will differ by
Multiples of π
So = -.80 + kπ
#2. X- Intercept Method
Steps: 1. rewrite the equation so = to 0
2. put in calculator – narrow window to 1 complete period (sin,cos 0-2π)
3. solutions are where curve crosses the x axis in the period
ex. sin x = .33  sin x -.33 = 0
x = 2.80 x = .33 since it is the sin function it repeats itself every 2π so
write as: 2.80 + 2kπ and .33 + 2kπ
#3. Other Examples of how to solve
Ex. 4cos²x + sinx – 2 = 0
a) already = 0 so use x-intercept method
b) both sin/cos period = 2π, so see where crosses
x-axis during this period
x = 1.00 x = 2.14 x = 3.78 x = 5.65
so: 1.00 +2kπ, 2.14 + 2kπ, 3.78 + 2kπ, 5.65 + 2kπ
All Example depend on knowing the period of a function**
Steps: 1)
2)
3)
4)
5)
write the equation in the form f(x) = 0
determine the period of f
graph over interval of the length of period
Use the calc’s zero finder to determine the x-intercepts
For each x intercept of u, all of the #’s are solutions of the equation
Ex. Solve tan x = -4cos2x
8.2 Inverse Trigonometric Functions – method used for solving trig equations w/o graph
Remember from Inverse Functions:
**a function cannot have an inverse unless its graph has the following property:
1) A horizontal line cannot intersect the graph more than once.
**ALL GRAPHS OF TRIG. FUNCTIONS WOULD FAIL THIS PROPERTY**
 unless we restrict their domains --- then an inverse can exist
#1 Inverse Sine/ Arcsine Function –
** If domain of Sine curve is restricted to [-π/2, π/2] – then it passes the HLT and has an
Inverse ** -- Inverse is called the Inverse sine or Arcsine
a) Arcsine is denoted by: g(x) = sin-1x or g(x) = arcsin x
b) Arcsine has a range of [-π/2, π/2] and a domain of [-1,1]
c) Formal definition is: for each #v in the interval [-1,1] there is exactly 1 #u in
the interval [-π/2, π/2] such that sin u = v
What this all mean:
Sin-1½  is saying: what radian in the interval [-π/2,π/2] gives a sine value of ½
=
Sin-1√3/2  is saying: what radian in the interval [-π/2,π/2] gives a sine value of
√3/2 =
Properties of Arcsine: (deals with compositions)
1) sin-1(sin u) = u
if -π/2≤u≤π/2
2) sin(sin-1v) = v
if -1≤v≤1
 2 step process: 1st find the sin of π/6 = ½
then think – what is the radian
value that gives sin = ½
** or see that automatically matches property 1 so answer is u (π/6)
ex. sin-1(sin π/6)
Ex. sin-1(7π/4)
 sin 7π/4 = -√2/2
But 7π/4 is not in our interval [-π/2,π/2] but –π/4
is, so the answer is –π/4
#2. Inverse Cosine/Arccosine Function
**If domain is restricted to interval [0,π] then passes the HLT and has an inverse – called
Inverse Cosine or Arccosine
a)Arccosine is denoted as g(x) = cos-1x or g(x)=arccos x
b) Arccosine has a domain = [-1,1] and a range = [0,π]
c) Formal definition: for each v in -1≤v≤1, cos-1v is the unique # u in the
interval [0,π] whose cosine is v – that is cos-1v = u exactly
when cosu = v
What this all means:
Cos-11  is saying what radian value in interval [0,π] gives a cosine value 1
Cos-1½is saying what radian value in interval [0,π] gives cosine value ½
Properties of Arccosine: (deals with compositions)
1) cos-1(cos u) = u
if 0≤u≤π
2) cos(cos-1v) = v
if -1≤v≤1
Ex. cos-1(.14) – press in calculator to get answer
Ex. cos-1(sin 4π/3)
#3. Inverse Tangent/Arctan function
**If original domain is restricted on interval (-π/2, π/2) –not included b/c asymptotes –
then passes HLT and has an inverse – called Inverse Tangent/Arctan Function
a) Arctan is denoted g(x) = tan-1x or g(x) = arctan x
b) Arctan has domain all real #’s and range of (-π/2,π/2)
c) Formal definition: for each real # v, tan-1v is the unique #u in the interval
(-π/2,π/2) whose tan is v – that is tan-1v=u exactly when
tan u = v
What it says:
Tan-11  what radian value on the interval (-π/2,π/2) gives you 1 (sin/cos)
Hint: know in Q1 b/c it is +
Tan-1136  press on calc (make sure in radian mode)
Properties of Arctan (deal with composition)
1) tan-1(tan u) = u if –π/2<u<π/2
2) tan(tan-1v) = v for every real #v
Ex. tan-1[tan(-4π/3)]
 tan -4π/3 = √3 -- what in interval gives √3 = π/3
Ex. find the exact value of sin(tan-1½)
a) let tan-1½ = u --- by property #1 tan u =½ and –π/2≤u≤π/2
** since u is positive it must be between 0 and π/2
b) use Pythagorean theorem tan = o/a
1² + 2² = c² --- c = √5
c) therefore sin(tan-1½) = sin = o/h = 1/√5
Ex. cos[sin-1(√3/5)]
8.3 Algebraic Solutions of Trigonometric Equations
**Now learn how to use algebra to solve trig. equations
Ex. Solving Basic Cosine Equations
Ex. cos x = .6
Steps: 1) use the inverse cosine function to find a solution:
2) another solution is found by finding the identity cos(-x) = cos (x)
use x from above = .9273
cos(-.9273) = cos(.9273) = .6
3) therefore the solutions are: -.9273 + 2kπ and .9273 + 2kπ (b/c period =2π)
Ex. Solving Basic Sine Equations
Ex. sin x = -.75
Steps: 1) do the inverse like above sin-1(x) where x is the # set = to
Sin(-.75) = -.8481
2) 2nd solution by identity sin(π-x) = sin x
Where x is the # from above: sin[π - -.8481] = sin 3.9897 = -.75
3) therefore solutions are: -.8481 + 2kπ and 3.9897 + 2kπ
Ex. Solving Basic Tangent Equations
ex. tanx = 3
Steps: 1) find inverse like above: tan-1(3) = 1.2490
2) tan will only have 1 solution on period so  1.2490 + kπ
Summary Chart: p 540
Equation
Possible values of c
Solutions
Sin x =c
-1<c<1
1. x=sin-1c + 2kπ
2. x =(π - sin-1c) + 2kπ
c=1
1. x = π/2 + 2kπ
c = -1
1. x = π/2 + 2kπ
c>1 or c<-1
1. No solution
c=0
1. kπ
------------------------------------------------------------------------------------cos x = c
-1<c<1
1. x = cos-1c + 2kπ
2. x = -cos-1c + 2kπ
c =1
1. x = 0 + 2kπ = 2kπ
c = -1
1. x = π + 2kπ
c>1 or c<-1
1. No solution
c=0
1. π/2 + kπ
---------------------------------------------------------------------------------------tan x = c
all real #’s
1. x = tan-1c + kπ
Using Solution Algorithm:
Ex. solve tan x = -4
Steps: tan-1(-4) = -1.3258
Solution x = -13258 + kπ
Ex. 3cosx + 1 = 0
Steps: get only trig. function on 1 side of equation cosx = -1/3
Ex. Solve sin u = ½ exactly without using a calculator
Steps:
Using Substitution and Basic Equation:
Ex. solve sin 4x = √3/2 exactly w/o using a calculator
1. u=4x
Algebraic Techniques – such as factoring/quad formula/basic identities
Ex. find the solution of 4sin²x + 5sinx + 1 = 0 in the interval [-π,π]
Ex. solve cosxsin²x = cosx  cosxsin²x – cosx = 0
Solve: Using Identities and Factoring
-6sin²x + cosx + 5 = 0
1) substitute 1-cos²x for sin²x to rewrite the eq. In
terms of cosine
-6(1-cos²x) + cosx + 5 = 0
-6 + 6cos²x + cosx + 5 = 0
6cos²x + cosx -1 = 0
6u² + u -1 = 0
(3u-1)(2u+1)=0
U = 1/3
u = -1/2
Cos x = 1/3
Cos-1(1/3) = 1.2310 + 2kπ
-cos-1(1/3) = -1.2310 + 2kπ
Cos-1(1/2) = 2π/3 + 2kπ
-cos-1(1/2) = -2π/3 + 2kπ
Solve Identities and Quad Formula
sec²x – 2tanx = 3
1) rewrite equation in terms of the tan function by putting
1 + tan²x for sec²x and use quad formula
(1+tan²x) – 2tanx-3 = 0
tan²x – 2tanx -2 =0
a=1
b=-2
c=-2