Solving Equations
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Trigonometry Lesson 10
Part One - Graphically Solving Equations
Solving trigonometric equations graphically:
When a question asks you to solve a system of trigonometric equations, they are looking for the
values of θ that make both equations true. There are two ways you can solve for θ : graphically
in your TI-83, and algebraically. Part I will show the graphing method, and Parts II & III will
focus on algebraic methods.
Example 1: Solve cosθ =
1
and state the general solutions:
2
Now use 2nd Æ Trace Æ Intersect to find the
points of intersection. They occur at 60º & 300º
In your TI-83, graph each
equation in degree mode.
If you extend the window, you will see that the
intersection points are in the same relative places,
one period later.
The first general solution is:
60º ± n(360º) or
and the second is:
300º ± n(360º) or
Example 2: Solve cos2θ =
π
3
± n(2π )
5π
± n(2π )
3
2
and state the general solutions:
2
Graph both equations in your TI-83,
then solve for the first two intersection points.
The first two intersection points are at 22.5º
and 157.5º. As you can see in the graph, the
solutions repeat themselves every period.
Since the b-value is 2, the period is 180º, or π.
The first general solution is:
22.5º ± n(180º) or
π
And the second is:
157.5º ± n(180º) or
8
± nπ
7π
± nπ
8
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Trigonometry Lesson 10
Part One - Graphically Solving Equations
Example 3: Graphically find the general solutions for 2sinθ - 3 = 0
Graph the two equations in your TI-83 and
solve by finding the points of intersection.
π
± n(2π )
3
2π
120º ± n(360º) or
± n(2π )
3
60º ± n(360º) or
(In this case, another method would be to find
the x-intercepts using 2nd Æ Trace Æ Zero)
Note that even if you manipulate the equation, you can still solve by graphing:
If you re-arrange the equation to 2sinθ =
3 by taking
3 to the other side, we get:
Solving, we still have the same answers of 60º & 120º
If we manipulate the equation again by dividing both sides by 2, we get: sinθ =
3
.
2
Solving this:
Once again, we still get the same answers of 60º & 120º
Manipulating an equation does NOT change the solution!
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Trigonometry Lesson 10
Part One - Graphically Solving Equations
Find the general solution (In degrees & radian fractions) for each of the following equations:
1) sin 3x =
3
2
2) cos 2 x = 1
1
2
3) sin 2 x = 0
4) sin 4 x = −
5) tan 2 x = 3
6) sin 2 x − 0.25 = 0
7) tan x + 3 = 0
8) sin
1
3
x=−
2
2
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Trigonometry Lesson 10
Part One - Graphically Solving Equations
π
⎛ 2π ⎞
± n⎜
⎟
9
⎝ 3 ⎠
2π
⎛ 2π ⎞
± n⎜
⎟
9
⎝ 3 ⎠
20D ± n(120D )
1)
40D ± n(120D )
± nπ
2) ± n(180D )
3) ± n(90 )
D
30D ± n(180D )
6)
150D ± n(180D )
7) 120D ± n(180D )
nπ
±
2
480D ± n(720D )
8)
600D ± n(720D )
± nπ
6
5π
± nπ
6
2π
± nπ
3
8π
± n ( 4π )
3
10π
± n ( 4π )
3
7π nπ
±
24 2
11π nπ
±
24
2
52.5D ± n(90D )
4)
82.5 ± n(90D )
5) 30D ± n(90D )
π
π
⎛π ⎞
± n⎜ ⎟
6
⎝2⎠
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Trigonometry Lesson 10
Part Two - Linear Equations
In this lesson, we will algebraically solve trigonometric equations. There are two main types
of equations you will be asked to solve: linear & nonlinear.
Note that you will still be able to solve all of these in your calculator as you did in the
previous lesson, but on the diploma they frequently have written response questions where
you need to present an algebraic solution.
Linear Trigonometric Equations:
Example 1: Solve: 4cosx + 2 = 0 in the domain 0 ≤ x < 2π
To solve this, we must get cosx by itself on the left side of the equation.
4cosx + 2 = 0
4cosx = -2
-2
cosx =
4
1
cosx = 2
From the unit circle, we know cosx is -
1
2π
4π
when x =
&
3
2
3
Example 2: Solve: 2sinxcosx = sinx
in the domain 0 ≤ x < 2π
We need to bring everything over to
the left side, then factor.
2sinxcosx - sinx = 0
sinx(2cosx -1) = 0
Now set each factor on the
left side equal to zero:
sinx = 0
x = 0, π
We don’t include 2π
since the domain is
using a < sign,
not a ≤ sign.
You may be tempted to divide
both sides of the equation by sinx
to cancel it out. Don’t do this!
In math, you are never allowed to
cancel variables on opposite sides
of the equation.
Suppose you made this error and
canceled sinx. Then you would get:
2sinxcosx = sinx
2cosx -1= 0
2cosx =1
1
2
π 5π
x= ,
3 3
cosx =
2cosx =1
1
cosx =
2
π 5π
x= ,
3 3
As you can see, we lose solutions
of 0 & π doing it this way.
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Trigonometry Lesson 10
Part Two - Linear Equations
Example 3: Solve:
sinx 1 sinx
+
=
in the domain 0 ≤ x < 2π
4
12
3
⎛ sinx 1 ⎞
⎛ sinx ⎞
12 ⎜
+ ⎟ =12 ⎜
⎟
12 ⎠
⎝ 4
⎝ 3 ⎠
3sinx +1= 4sinx
Multiply both sides by the
common denominator, which
is 12. This will eliminate the
fractions.
1= sinx
x=
π
2
Example 4: Solve sinxsecxcotx = sinxsecx in the domain 0 ≤ x < 2π
Bring all terms to one
side so you can factor.
sinxsecxcotx - sinxsecx = 0
sinxsecx(cotx -1) = 0
sinx = 0
x = 0,π
secx = 0
1
=0
cosx
x = No solutions
cotx -1= 0
cotx =1
π 5π
x= ,
4 4
The combined solution set is:
π
5π
0, , π ,
4
4
Example 5: Solve 3sinx = 2 in the domain 0 ≤ x < 2π
3sinx = 2
2
sinx =
3
2
is not on the unit circle, we are forced to do this
3
equation graphically in the calculator.
Since
41.8º = 0.73 rad
138.2º = 2.41 rad
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Trigonometry Lesson 10
Part Two - Linear Equations
Solve each of the following in the domain 0 ≤ x < 2π
1
2
1) sin x − = 0
2) 2 cos x − 3 = 0
3) 4sin x + 3 = 3sin x + 2
4) 2sin x cos x = cos x
5) 3 tan x = 3
6) 2 cos x + 1 = 2
7) sin x cos x tan x + sin x cos x = 0
8) 2 cos x(cos x + ) = 0
1
2
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Trigonometry Lesson 10
Part Two - Linear Equations
9) (sec x + 1)(sin x − 1)(cot x − 1) = 0
10) csc x cot x + cot x = 0
11) tan x cos x + cos x = 0
12) (cos x − 1)(tan x − 1) = 0
13) sec x cos x + sec x = 0
14)
sin x sin x
=
2
3
16)
csc x csc x 16
+
=
15
5
3
15)
tan x tan x −1
−
=
6
2
3
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Trigonometry Lesson 10
Part Two - Linear Equations
π 5π
1) x = ,
π 5π
12) x = 0, ,
6 6
4 4
π 11π
2) x = ,
6
3) x =
13)
6
3π
2
2sin x cos x − cos x = 0
4) cos x(2sin x − 1) = 0
x=
π π 5π 3π
, , ,
6 2 6 2
sec x(cos x + 1) = 0
x =π
3sin x = 2sin x
3sin x − 2sin x = 0
14)
sin x = 0
x = 0, π
⎛ tan x tan x ⎞
⎛ −1 ⎞
−
6⎜
⎟ = 6⎜ ⎟
3 ⎠
⎝ 2
⎝ 6 ⎠
3 tan x − 2 tan x = −1
15)
tan x = −1
tan x = 1
5)
π 5π
x= ,
4 4
x=
1
2
6)
π 5π
x= ,
3 3
cos x =
16)
sin x cos x(tan x + 1) = 0
7)
π 3π
3π 7π
x = 0, , , π , ,
2 4
2 4
π 2π 4π 3π
8) x = ,
2
3
π π
,
9) x = , , π ,
4 2
3π 7π
,
4 4
3
,
2
5π
4
⎛ csc x csc x ⎞
⎛ 16 ⎞
15 ⎜
+
⎟ = 15 ⎜ ⎟
3 ⎠
⎝ 5
⎝ 15 ⎠
3csc x + 5csc x = 16
8csc x = 16
csc x = 2
1
2
π 5π
x= ,
6 6
sin x =
cot x(csc x + 1) = 0
10)
π 3π
x= ,
2 2
cos x(tan x + 1) = 0
11)
π 3π 3π 7π
x= , , ,
2 4 2 4
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Trigonometry Lesson 10
Part Three - Nonlinear Equations
Quadratic Trigonometric Equations:
Example 1: Solve 4sin 2 x - 3 = 0 in the domain 0 ≤ x < 2π
4sin 2 x = 3
3
sin 2 x =
4
3
sin 2 x =
4
3
2
π 2π 4π 5π
x= ,
,
,
3 3 3 3
sinx = ±
Example 2: Solve cos 2 x - cosx = 0 in the domain 0 ≤ x < 2π
cos 2 x - cosx = 0
cosx(cosx -1) = 0
cosx = 0
π 3π
x= ,
2 2
cosx -1= 0
cosx =1
x=0
π 3π
The complete solution is: x = 0, ,
2 2
Example 3: Solve 2cos 2 x = 3cosx -1 in the domain 0 ≤ x < 2π
2cos 2 x - 3cosx +1= 0
(2cosx -1)(cosx -1) = 0
2cosx -1= 0
2cosx =1
cosx -1= 0
1
2
π 5π
x= ,
3 3
cosx =1
x=0
cosx =
π 5π
The complete solution is: x = 0, ,
3 3
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Trigonometry Lesson 10
Part Three - Nonlinear Equations
Example 4: Solve sin 3 x - 5sin 2 x +6sinx = 0 in the domain 0 ≤ x < 2π
sinx(sin 2x - 5sinx + 6) = 0
sinx(sinx - 2)(sinx - 3) = 0
sinx - 2 = 0
sinx - 3 = 0
π 3π
sinx = 2
sinx = 3
2 2
x = no solution
x = no solution
sinx = 0
x=
,
Example 5: Solve tan8 x - tan 4 x = 0 in the domain 0 ≤ x < 2π , and state the
general solution
4
Watch out for
difference of squares!
4
tan x(tan x -1) = 0
tan 4 x(tan 2 x +1)(tan2 x -1) = 0
tan 4 x(tan 2 x +1)(tanx +1)(tanx -1) = 0
tan 4 x = 0
tan 2 x +1= 0
4
tan 2 x = -1
tanx +1= 0
tanx = -1
No solution
x=
tan 4 x = 4 0
tanx = 0
x = 0,π
3π 7π
,
4 4
The complete solution set for 0 ≤ x < 2π is: x = 0,
The general solution is: x = ±nπ and x =
tanx -1= 0
tanx =1
π
±n
x=
π 3π
,
4 4
,π ,
π 5π
,
4 4
5π 7π
,
4 4
π
4
2
These two general solutions will account for all the angles we found.
Example 6: Solve 3cot 2 x - cotx -1= 0 in the domain 0 ≤ x < 2π
This equation cannot be factored, so graph and find the solutions in radian
decimals:
0.92 rad, 1.98 rad, 4.06 rad, 5.12 rad
Graph in your TI-83 as:
3
2
(tan(x))
-
1
tan(x)
-1
Then use 2nd Æ Trace Æ Zero
to find x-intercepts.
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Trigonometry Lesson 10
Part Three - Nonlinear Equations
Solve each of the following in the domain 0 ≤ x < 2π
1) cos 2 x =
3
4
7) 6 cos 2 x − 3cos x − 3 = 0
8) 2sin 2 x − 3sin x + 1 = 0
1
2) sin x − = 0
4
2
9) 4 cos 2 x + 2 cos x − 2 = 0
3) 3 tan x = 3
2
4) 4sin 2 x − 3 = 0
10) 2 cos3 x + cos2 x − cos x = 0
5) 2sin 2 x − sin x − 1 = 0
11) tan 4 x − tan 2 x = 0
6) 2 sin 2 x = 1 − sin x
12) cos8 x − cos 4 x = 0
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Trigonometry Lesson 10
Part Three - Nonlinear Equations
ANSWERS:
π 5π 7π 11π
1) x = ,
6 6
,
6
,
6
π 5π 7π 11π
2) x = ,
6 6
,
6
,
6
2(2 cos 2 x + cos x − 1) = 0
9) 2(2 cos x − 1)(cos x + 1) = 0
x=
π
3
,π ,
5π
3
π 3π 5π 7π
3) x = ,
4 4
,
4
,
4
π 2π 4π 5π
4) x = ,
3
3
,
3
,
3
(2sin x + 1)(sin x − 1) = 0
5)
π 7π 11π
x= ,
,
2 6 6
cos x(2 cos 2 x + cos x − 1) = 0
10) cos x(2 cos x − 1)(cos x + 1) = 0
x=
π π
,
3 2
,π ,
3π 5π
,
2 3
tan 2 x(tan 2 x − 1) = 0
11) tan 2 x(tan x + 1)(tan x − 1) = 0
x = 0,
π 3π
(2sin x − 1)(sin x + 1) = 0
6)
π 5π 3π
x= , ,
6 6 2
,
4 4
,π ,
5π 7π
,
4 4
cos 4 x(cos 4 x − 1) = 0
cos 4 x(cos 2 x + 1)(cos 2 x − 1) = 0
3(2 cos x − cos x − 1) = 0
7) 3(2 cos x + 1)(cos x − 1) = 0
2π 4π
x = 0,
,
3 3
2
12) cos 4 x(cos 2 x + 1)(cos x + 1)(cos x − 1) = 0
x = 0,
π
2
,π ,
3π
2
(2sin x − 1)(sin x − 1) = 0
8)
π π 5π
x= , ,
6 2 6
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Trigonometry Lesson 10
Part Four - Algebraically Solving Multiple Angles
Algebraically Solving Double, Triple, and Half Angles:
While technology can be used to solve equations involving double & triple angles, it is
advantageous to understand the algebraic process involved with these question types.
Example 1: Solve sin2θ =
3
algebraically over the interval 0 ≤ x ≤ 2π .
2
To solve this equation algebraically, you need to perform the following steps:
3
. We can do this easily using the unit
2
π
2π
and
.
circle. The answer to this equation is x =
3
3
Step 1) Start by solving the equation sin x =
Step 2) Add 2π to each of the angles we found.
π
6π 7π
=
3
3 3
3
2π
2π 6π 8π
+ 2π =
+
=
3
3
3
3
+ 2π =
π
Step 3) Finally, take all your solutions
The answer is
+
Don’t forget that adding fractions
requires a common denominator.
π 2π 7π 8π
3
,
3
,
3
,
π π 7π 4π
, , ,
6 3 6 3
3
and divide by 2. (Or multiply by ½)
The general solution is:
You can verify the
results by graphing
and checking the
x-coordinates
of the points of
intersection.
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Trigonometry Lesson 10
Part Four - Algebraically Solving Multiple Angles
2
for the domain 0 ≤ θ ≤ 2π
2
2
Step 1) Start by solving the equation cos x =
. We can do this using the unit circle.
2
π
7π
and
The answer to this equation is x =
.
4
4
Example 2: Solve cos3θ =
Step 2) Add 2π to each of the angles we found.
π
8π 9π
=
4
4 4
4
7π
7π 8π 15π
+ 2π =
+
=
4
4
4
4
+ 2π =
π
+
Step 3) Add 2π to each of the angles we found in Step 2.
9π
9π 8π 17π
+ 2π =
+
=
4
4
4
4
15π
15π 8π 23π
+ 2π =
+
=
4
4
4
4
Step 4) Finally, take all your solutions
The answer is
π 7π 9π 15π 17π 23π
,
4
,
4
,
4
,
4
,
4
4
and divide by 3. (Or multiply by 1/3)
π 7π 3π 5π 17π 23π
, , , ,
,
12 12 4 4 12 12
The general solution is:
You can verify the
results by graphing
and checking the
x-coordinates of the
points of intersection.
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Trigonometry Lesson 10
Part Four - Algebraically Solving Multiple Angles
1
3
Example 3: Solve sin θ =
for the domain 0 ≤ θ ≤ 2π
2
2
To solve this equation algebraically, you need to perform the following steps:
Step 1) Start by solving the equation sin x =
The answer to this equation is x =
Step 2) Divide each angle by
π
3
and
2π
.
3
3
. Do this using the unit circle.
2
1
2π
4π
and
(Or multiply by 2) to obtain
.
2
3
3
The general solution is:
You can verify the
results by
graphing and
checking the
x-coordinates of
the points of
intersection
Questions: Algebraically solve for θ over the interval 0 ≤ θ ≤ 2 π
3
2
1) sin 2θ = −
4) sin 2θ = −1
Answers:
2) cos 2θ =
3
2
3) cos 2θ = −
1
2
5) cos 3θ =
3
2
2
⎛1 ⎞
6) cos ⎜ θ ⎟ =
⎝2 ⎠ 2
2π 5π 5π 11π
, , ,
3 6 3 6
π 11π 13π 23π
,
,
2) θ = ,
12 12 12 12
π 2π 4π 5π
,
,
3) θ = ,
3 3 3 3
3π 7π
,
4) θ =
4 4
1) θ =
5) θ = π
11π 13π 23π 25π 35π
,
,
,
,
,
18 18 18 18 18 18
6) θ =
π
2
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