1. 8.5 Solving More Difficult Trig Equations

2. 8.5 Solving Trig Equations
Objectives:
Solve basic trig equations
Use factoring and trig identities to solve complicated trig equations
Vocabulary: sine, cosine, tangent, cosecant, secant, cotangent, cofunction

3. 8.5 Solving Trig Equations
WARNING!!!!!
Do not divide by sin, cos, tan, sec, csc or cot! You may lose answers.
Always check your solutions

4. 8.5 Solving Trig Equations Cont
Objective to solve simple trigonometric equations and to apply them.

5. SOLVE: 3 1 =2𝑠𝑖𝑛𝜃 3 2 =𝑠𝑖𝑛𝜃 𝑠𝑖𝑛 𝜋 3 = 3 2 𝜃= 𝜋 3 , 2𝜋 3
1. Isolate the trig functions using algebra and identities
3 1 =2𝑠𝑖𝑛𝜃
3 2 =𝑠𝑖𝑛𝜃
2. Determine the quadrants
Sin + in first and second
𝑠𝑖𝑛 𝜋 3 =
3. Find the angle measure
4. Use reference angles to determine solutions
𝜃= 𝜋 3 , 2𝜋 3

6. Solve cos θ + 1 = sin θ in [0, 2]
𝑆𝑖𝑛𝑐𝑒 𝑠𝑖𝑛 2 + 𝑐𝑜𝑠 2 =1
𝑠𝑖𝑛= 1 − 𝑐𝑜𝑠 2
Like solving algebraic equation, once we squared the original trigonometric equation, it may generate some extraneous solution. We need to check the solutions.
The only solutions are /2 and .

7. If the values that these trig functions equal are NOT exact values on the unit circle you will need to use your calculator.
from calculator:
this value is somewhere in Quadrant I
What other quadrant would have the same cosine value (same x value on the unit circle)?
Quadrant IV
This angle is 2 minus angle from calculator.

8. Solve tan2x – 3tanx – 4 = 0 in 0 ≤ x < 360.
tan x – 4 = 0 or tan x + 1 = 0
tan x = 4 or tan x = –1
x = arctan(4) or x = arctan(-1)
x = 76° or x = 135°
Quad check?
x = 256° or x = 315°

9. 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
 = arcsin( )  and  = π  arcsin( )  1 3

10. Solve. 3sinx + 4 = -1/sinx in 0 ≤ x ≤ 2 .
Let x = sin x, then
3x + 4 = -1/x
x(3x + 4) = x(-1/x)
3x2 + 4x = -1
(3x + 1)(x + 1) = 0
3x + 1= 0 or x + 1 = 0
x = -1/3 or x = –1
sin x = -1/3 or sin x = –1
x =  + arcsin(1/3) or x = 3/2
x = 2 – arcsin(1/3)

11. Solve 5cos2 + cos  – 3 = 0 for 0 ≤  ≤ π.
The equation is quadratic. Let u = cos and solve
5u2 + u  3 = 0.
u = -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( ) =

12. Since we squared the original equation we have to check our answer.

13. Since we squared the original equation we have to check our answer.

14. Inclination and Slope
The inclination of a line is the angle , where 0   <  (0o   < 180o), that is measured from the positive x-axis to the line.
The line at the left above has positive slope and the right one has negative slope. How does the slope connect to the trigonometric functions?

15. Since line l2 is parallel to line l1, thus,
Suppose that the inclination of line l2 is , the line l1 is passes through the origin and is parallel to l2 . So the inclination of line l1 is . Let P be the point where l1 intersects the unit circle. Then P (cos, sin). The slope of line l1 can be calculated between point P and the origin: y
P (cos, sin)
r = 1   x l2 l1
Since line l2 is parallel to line l1, thus,

16. Example Find the inclination of line 2x + 5y = 15
[Solution] Rewrite the equation as y = –2/5x + 3
Then the slope is m = –2/5 = tan
Therefore  = arctan (–2/5)  –21.8o
The reference angle is  = arctan (2/5)  21.8o.
The inclination is  – arctan(2/5) = 180o – 21.8o  158.2o