Transcript Link to ppt Lesson Notes - Mr Santowski`s Math Page

Lesson 53 – Solving
Trigonometric Equations
Math 2 Honors - Santowski
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Math 2 Honors - Santowski
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(A) Review - Solving Equations

The general strategy to ALGEBRAICALLY
solving ANY equations is to:
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(a) Isolate the “base” function containing the
unknown

(b) Isolate the unknown by using the inverse of
the base function
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(A) Review - Solving Equations

And a second alternative to solving
ALGEBRAICALLY is to solve GRAPHICALLY by
either:

(a) looking for an intersection point for f(x) = g(x)

(b) looking for the zeroes/roots of a rearranged
eqn in the form of f(x) – g(x) = 0
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(A) Examples


Here are some easy examples to start with
Let’s use the domain -2π < x < 2π and then
work to an infinite domain
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(a) sin( x)  1  0
(b)
(c) tan( x)  1  0
1
(e) sin( 2 x) 
2
(d) 2 sin( x)  3  0
(f)
3 sec( x)  2  0
2 csc( 4 x)  2
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(A) Examples - Graphically
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(B) Examples – With Calculator

Solve the equation 3sin(x) – 2 = 0

We can rearrange as sin(x) = 2/3 so x = sin-1(2/3) giving
us 41.8° (and the second angle being 180° - 41.8° =
138.2°)

Note that the ratio 2/3 is not one of our standard ratios
corresponding to our “standard” angles (30,45,60), so we
would use a calculator to actually find the related acute
angle of 41.8°
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(B) Examples – With Calculator

We can now solve the equation 3sin(x) – 2 = 0 by graphing f(x) = 3sin(x) – 2
and looking for the x-intercepts
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(B) Examples – With Calculator




Notice that there are 2 solutions

within the limited domain of 0° < 
< 360°
However, if we expand our
domain, then we get two new
solutions for every additional
period we add
The new solutions are related to
the original solutions, as they
represent the positive and negative
co-terminal angles
We can determine their values by
simply adding or subtracting
multiples of 360° (the period of the
given function)
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(B) Examples – With Calculator


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

Solve 4tan(2x) + 3 = 2
Again, we can set it up algebraically as tan(2x) = -1/4 and thus (2x)
= tan-1(-1/4) so x = ½ tan-1(-1/4)
So thus x = ½ of 166° = 84°
and x = ½ of 346° = 173°
To set it up graphically, we will make one minor change: we have
two graphing options  we can graph f(x) = 4tan(2x) + 1 and find
the x-intercepts OR we can graph f(x) = 4tan(2x) + 3 and find out
where f(x) = 2  to do this, we will simply graph y = 2 as a second
equation and find out where the two graphs intersect
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(B) Examples – With Calculator
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(B) Examples – With Calculator




Notice that there are 2 solutions within the
limited domain of 0° <  < 180° (NOTE:
the period of a regular tan function is 180°)
However, if we expand our domain, then
we get two new solutions for every
additional period we add
The new solutions are related to the
original solutions, as they represent the
positive and negative co-terminal angles
We can determine their values by simply
adding or subtracting multiples of 90° (the
period of the given function  the effect of
the 2x in tan(2x) is a horizontal
compression by a factor of 2, so we have
reduced the period by a factor of 2)
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(C) Quadratic Trigonometric Equations

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
Quadratic trig eqns contain terms like sin2(), cos2()
Recall the Pythagorean identity (sin2() + cos2() = 1)
Recall how to factor simple trinomials like x2 + 2x – 35 =
0  (x + 7)(x – 5) = 0
Recall how to factor difference of square trinomials like
4x2 – 25 = 0  (2x – 5)(2x + 5) = 0
Recall how to factor trinomials in the form of 3x2 – x – 4
= 0  using decomposition or guess & check  (3x 4)(x + 1) = 0
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(C) Quadratic Trigonometric Equations

Solve 2cos2() = 1 if 0° <  < 360 °
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(C) Quadratic Trigonometric Equations

Solve 2cos2() = 1 if 0° <  < 360 °
2 cos ( )  1
2
cos ( )  1 2
cos( )   1
2
2
 1 
   cos  
2
  45 ,135 ,225 ,315
1
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(C) Quadratic Trigonometric Equations

Solve cos2(x) + 2cos(x) = 0 for 0 < x < 2
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(C) Quadratic Trigonometric Equations

Solve cos2(x) + 2cos(x) = 0 for 0 < x < 2
cos2 ( x )  2 cos( x )  0
cos( x )  cos( x )  2  0
(i )  cos( x )  0
x  cos 1 ( 0)
 3
x
,
2 2
(ii )  cos( x )   2
x  cos 1 (  2)
xR
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(C) Quadratic Trigonometric Equations

Solve 2cos2(x) - 3cos(x) + 1 = 0 for 0 < x < 2
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(C) Quadratic Trigonometric Equations

Solve 2cos2(x) - 3cos(x) + 1 = 0 for 0 < x < 2
2 cos2 ( x )  3 cos( x )  1  0
(2 cos( x )  1)(cos( x )  1)  0
(i )  2 cos( x )  1  0
cos( x )  1 2
x   3 , 5 3
(ii )  cos( x )  1
x  0,2
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(C) Quadratic Trigonometric Equations
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Solve 2cos2(x) - sin(x) - 1 = 0
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(C) Quadratic Trigonometric Equations
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Solve 2cos2(x) - sin(x) - 1 = 0  notice we have both
sin(x) and cos(x) in the equation  so use the
Pythagorean identity to make changes
The equation becomes 2(1 – sin2(x)) – sin(x) – 1 = 0
So this is now –2sin2(x) – sin(x) + 1=0 or we can make it
0 = 2sin2(x) + sin(x) – 1 which we can now factor and
solve on 0° < x < 360 °
(2sin(x) - 1)(sin(x) + 1) = 0
So 2sin(x) - 1 = 0 °, meaning x = 30 ° and 150 °
And sin(x) + 1 = 0 °, meaning x = 270 °
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(D) Other Examples

Solve for x:
(a) sin( x)  3 cos( x)  0
(b) sin( 2 x)  sin( x)

3

(c) cos 3x   
6 2

(d ) 4 sin 2 ( x)  3
(e) 2 sec 2 ( x)  1
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Homework
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HW
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S14.6, p926,
Q9,13,17,19,21,23,31,35,37,39,41,45,49,50
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