Post on 27-Oct-2014
8.5 Solving More Difficult Trigonometric Equations
Objective To use trigonometric identities or technology to solve more difficult trigonometric equations.
x
y
[Solution] Take the fourth root of both sides to obtain:
cos(2 )=±2
3
From the unit circle, the solutions for 2 are 2 = ± + kπ, k any integer.
π 6
Example 1: To find all solutions of cos4(2 ) = .9
16
1 π 6-π 6
x = -2
3 x =2
3
ππ
Answer: = ± + k , for k is any integer.12
2
Example 2: Find all solutions of the trigonometric equation: tan2 + tan = 0.
The solutions for tan = 0 are the values = kπ, for k is any integer.
Therefore, tan = 0 or tan = -1.
tan2 + tan = 0 Original equation
tan (tan +1) = 0 Factor.
The solutions for tan = -1 are = - + kπ, for k is any integer.
π 4
Some trigonometric equations can be transformed into equations that have quadratic form.
Example 3: The trigonometric equation
2 cos2 – 3 sin – 3 = 0 .
2 sin2 + 3 sin + 1 = 0 implies that
= -π/2 + 2kπ, from sin = -1
(2 sin + 1)(sin + 1) = 0.
Therefore, 2 sin + 1 = 0 or sin + 1 = 0.
Solutions: = - + 2kπ and = + 2kπ, from sin = -π
67π 6
1 2
It follows that sin = - or sin = -1.1 2
[Solution] Use the Pythagorean identity:
2 (1 – sin2 ) – 3 sin – 3 = 0 .
Steps for Solving:
• Isolate the Trigonometric function.• Then solve for the angle exactly if the ratio
represents known special triangle ratios.• Or solve for the angle approximately using the
appropriate inverse trigonometric function.
Complete the List of Solutions:
If you are not restricted to a specific interval and are asked to give a complete list of solutions (general solution), then remember that adding on any integer multiple of 2π represents a co-terminal angle with the equivalent trigonometric ratio.
8 sin = 3(1 sin2 ) Use the Pythagorean Identity.
[Solution] Rewrite the equation in terms of only one trigonometric function.
Example 4: Solve 8 sin = 3 cos2 with in the interval
[0, 2π].
3 sin2 + 8 sin 3 = 0. A “quadratic” equation with sin x as the variable
Therefore, 3 sin 1 = 0 or sin + 3 = 0
(3 sin 1)(sin + 3) = 0 Factor.
= arcsin( ) 0.3398 and = π arcsin( ) 2.8107.1 3
1 3
Solutions: sin = or sin = -31 3
Example 5: 5cos2 + cos – 3 = 0 for 0 ≤ ≤ π.
This is the range of the inverse cosine function.
The solutions are: = cos 1(0.6810249 ) = 0.8216349 and = cos 1(0.8810249) = 2.6488206
Therefore, cos = 0.6810249 or –0.8810249.
Use the calculator to find values of in 0 ≤ ≤ π.
[Solution] The equation is quadratic. Let u = cos and solve 5u2 + u 3 = 0.
u = -1 ± = 0.6810249 or -0.881024961 10
Example 6: Solve a trigonometric equation by factoring.
tan x sin x sin x
tan x sin x sin x 0
sin x tan x 1 0
tan 1 0 or sin 0x x
245
2 24 4
(2 1)4
x k
x k k
k
[Solution]
tan 1 or sin 0x x
x k
4x k
Example 7: Solve tan2x – 3tanx – 4 = 0 in 0 ≤ x ≤ 2.
[Solution] Let u = tan x, then
x2 – 3x – 4 = 0(x – 4)(x + 1) = 0x – 4 = 0 or x + 1 = 0x = 4 or x = –1
tan x = 4 or tan x = –1 x = arctan4 or x = 3/4 (x = 76° or x = 135°)
x = + arctan4 or x = 7/4 ( x = 256° or x = 315°)
Example 8: Solve. 3sinx + 4 = 1/sinx in 0 ≤ x ≤ 2 .
[Solution] Let x = sin x, then
3x + 4 = 1/x
x(3x + 4) = x(1/x) 3x2 + 4x = 1(3x – 1)(x + 1) = 03x – 1= 0 or x + 1 = 0x = 1/3 or x = –1sin x = 1/3 or sin x = –1
x = arcsin(1/3) or x = 3/2( x = 19° or x = 270°)x = – arcsin(1/3)( x = 161°)
2 2 Solve sin sin cos for 0 2 .x x x x Example 9 :
2 2sin sin cosx x x
2 2sin sin 1 sinx x x 22sin sin 1 0x x
2sin 1 sin 1 0x x
2sin 1 0 sin 1 0x x 1
sin sin 12
x x
7 11,
6 6 2x x
[Solution]
Solve sin tan 3sin . x x xExample 10 :
sin tan 3sin x x x
sin tan 3sin =0x x x
sin tan 3 0x x
sin 0 tan 3 0x x
0,x
tan 3x
arctan 3, arctan 3x
[Solution]
Particular solutions are
2x karctan 3x k General solutions are
2x k x k
Example 11: Solve 2 cos x + sec x = 0
[Solution]
2
12cos 0
cos1
2cos 1 0cos
xx
xx
2 1cos ,
2x
1cos
2x
is not a real number,
thus the equation has no solution.
10
cos x
Therefore,22cos 1 0x
10
cos x
or
However, So, 22cos 1x
Example 12: Solve cos x + 1 = sin x in [0, 2]
[Solution]
3, or
2 2x x
2cos 1 = 1 cos
Since 1 cos 1, 0 cos 1 2 2Thus sin 0 sin 1 cos
2 2cos 2cos 1 1 cosx x 22cos 2cos 0x x
2cos (cos 1) 0x x
2cos 0 or cos 1 0x x
cos 0 or cos 1x x Since we squared the original equation we have to check our answer. The only solutions are /2 and .
Like solving algebraic equation, once we squared the original trigonometric equation, it may generate some extraneous solution. We need to check the solutions.
Book Example 5. Solve 2sin cos 1 for 0 < 360 .
2sin cos 1 2 2 2sin 1 cos sin 1 cos
22 1 cos cos 1
2
222 1 cos cos 1 2 24 1 cos cos 2cos 1
2 24 4cos cos 2cos 1 20 5cos 2cos 3
5cos 3 cos 1 0
5cos 3 0 cos 1 0 3
cos cos 15
53.3 ,306.9 180
Since we squared the original equation we have to check our answer.
2sin53.1 cos53.1 1 1.6 1.6
2sin180 cos180 1 0 0
2sin306.9 cos306.9 1 1.6 1.6
the only solutions are 53.1 and 180
Book Example 5. Solve 2sin cos 1 for 0 < 360 .
2sin cos 1 Since 1 cos 1, 0 cos 1 2
Then 2sin cos 1 0
2
222 1 cos cos 1 2 24 1 cos cos 2cos 1
2 24 4cos cos 2cos 1 20 5cos 2cos 3
5cos 3 cos 1 0
5cos 3 0 cos 1 0 3
cos cos 15
53.3 ,306.9 180
Since we squared the original equation we have to check our answer.
the only solutions are 53.1 and 180
2Thus sin 0 sin 1 cos
2sin53.1 cos53.1 1 1.6 1.6
2sin180 cos180 1 0 0
2sin306.9 cos306.9 1 1.6 1.6
a)
Reference Angle:
Therefore:
b)
The equation cannotbe factored. Therefore,use the quadratic equation to find the roots:
Reference Angles:
sin 13
tan b b ac
a
2 42
tan( ) ( ) ( )( )
( )
1 1 4 3 12 3
2
4 1sin sin 3 1sin
0 3398.
0 3398 2 8018. .and
3 1 02tan tan
tan . tan . 0 43 0 76or
0 4061. 0 6499.
2 7355 5 877 0 6499 3 7915. , . , . , .
Using a Graphing Calculator to Solve Trigonometric Equations0 2
y = 3sin -1
5.2.13
3 tan2 tan 1 0
Using a Graphing Calculator to Solve Trigonometric Equations
Therefore, and
Remark
Solving trigonometric equations is a long lasting task through out the entire trigonometry. After we learned the sum and difference of angles, double angles, triple angles, we will be able solve some much more difficult trigonometric equations.
Assignment
P. 326 #1 – 22