1. Solving Trigonometric Equations
Digital Lesson
Solving Trigonometric Equations

2. sin x = is a trigonometric equation.
x = is one of infinitely many solutions of y = sin x. π 6 -1 x y 1
-19π 6
-11π
-7π π 5π
13π
17π
25π
y = -π
-2π
-3π π 2π 3π 4π
All the solutions for x can be expressed in the form of a general solution.
x = k π and x = k π (k = 0, ±1, ± 2, ± 3,  ). 6 π
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y=sin x

3. Example: General Solution
Find the general solution for the equation sec  = 2.
From cos  = , it follows that cos  = . 1
sec 
cos( kπ) = π 3 -π x y Q 1 P
All values of  for which cos  = are solutions of the equation. Two solutions are  = ± All angles that are coterminal with ± are also solutions and can be expressed by adding integer multiples of 2π. π 3
The general solution can be written as
 = ± kπ . π 3
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Example: General Solution

4. Example: Solve tan x = 1.
The graph of y = 1 intersects the graph of y = tan x infinitely many times. y 2 x -π π 2π 3π
x = -3π
y = tan(x)
x = -π
x = π
x = 3π
x = 5π
y = 1
- π – 2π 4
- π – π
π + π
π + 2π
π + 3π
Points of intersection are at x = and every multiple of π added or subtracted from . π 4
General solution: x = + kπ for k any integer. π 4
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Example: Solve tan x=1

5. Example: Solve the Equation
Example: Solve the equation 3sin x = sin x for  ≤ x ≤ . π 2
2sin x = 0
Collect like terms.
3sin x  sin x = 0
3sin x = sin x
sin x =  1 x y
y = - 1 -π 4
x =  is the only solution in the interval  ≤ x ≤ . π 2 4
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Example: Solve the Equation

6. Example: Find all solutions using unit circle
Example: To find all solutions of cos4(2x) = 9 16
Take the fourth root of both sides to obtain: cos(2x)= ± x y
From the unit circle, the solutions for 2 are 2 = ± kπ, k any integer. π 6 π 1 π 6 -π π
x = -
x =
Answer:  = ± k ( ), for k any integer. 12 π 2
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Example: Find all solutions using unit circle

7. Example: Find all solutions
Find all solutions of the trigonometric equation: tan2  + tan  = 0.
tan2  + tan  = 0
Original equation
tan  (tan  +1) = 0
Factor.
Therefore, tan  = 0 or tan  = -1.
The solutions for tan  = 0 are the values  = kπ, for k any integer.
The solutions for tan  = 1 are  = kπ, for k any integer. π 4
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Example: Find all solutions

8. 2 sin2  + 3 sin  + 1 = 0 implies that
The trigonometric equation 2 sin2  + 3 sin  + 1 = 0 is quadratic in form.
2 sin2  + 3 sin  + 1 = 0 implies that
(2 sin  + 1)(sin  + 1) = 0.
Therefore, 2 sin  + 1 = 0 or sin  + 1 = 0.
It follows that sin  = - or sin  = -1. 1 2
Solutions:  = kπ and  = kπ, from sin  = - π 6 7π 1 2
 = -π + 2kπ, from sin  = -1
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Quadratic Form

9. Example: Solutions in an interval
Example: Solve 8 sin  = 3 cos2  with  in the interval [0, 2π].
Rewrite the equation in terms of only one trigonometric function.
8 sin  = 3(1 sin2  )
Use the Pythagorean Identity.
3 sin2  + 8 sin   3 = 0.
A “quadratic” equation with sin x as the variable
(3 sin   1)(sin  + 3) = 0
Factor.
Therefore, 3 sin   1 = 0 or sin  + 3 = 0
Solutions: sin  = or sin  = -3 1 3
 = sin1( ) = and  = π  sin1( ) = 1 3
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Example: Solutions in an interval

10. Example: Solve quadratic equation
Solve: 5cos2  + cos  – 3 = 0 for 0 ≤  ≤ π.
The equation is quadratic. Let y = cos  and solve 5y2 + y  3 = 0.
y = (-1 ± ) = or 10
Therefore, cos  = or –
Use the calculator to find values of  in 0 ≤  ≤ π.
This is the range of the inverse cosine function.
The solutions are:
 = cos 1( ) = and  = cos 1( ) =
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Example: Solve quadratic equation

11. Example: Find points of intersection
Example: Find the intersection points of the graphs of y = sin  and y = cos . x y π 4
+ kπ 1
The two solutions for  between 0 and 2π are and 5π 4 π 5 -π 4 π 4
+ kπ
The graphs of y = sin  and y = cos  intersect at points where sin  = cos . This is true only for triangles.
The general solution is  = + kπ, for k any integer. π 4
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Example: Find points of intersection