1. 10.4 Solve Trigonometric Equations

2. Solving trig equations… What’s the difference?

3. Solve a trig equation Solve 2 sin x – 3 = 0. SOLUTION
First isolate sin x on one side of the equation.
2 sin x – 3 =
Write original equation.
2 sin x 3 =
Add 3 to each side.
sin x 3 2 =
Divide each side by 2.
One solution of sin x = 3 2
in the interval 0 ≤ x < 2 π is 3 2 =
sin–1 π 3 . = x π 3 –
= π 2π 3 . =
The other solution in the interval is x

5. (where n is any integer)
Moreover, because y = sin x is periodic, there will be infinitely many solutions.
You can use the two solutions found above to write the general solution:
+ 2nπ π 3 =
+ 2nπ 2π 3 = x or x
(where n is any integer)
CHECK

6. Solve a trigonometric equation in an interval
Solve 9 tan2 x + 2 = 3 in the interval 0 ≤ x <2π.
9 tan2 x + 2 3 =
Write original equation.
9 tan2 x 1 =
Subtract 2 from each side. 1 9 =
tan2 x
Divide each side by 9. 1 3 + – =
Take square roots of each side.
tan x
Using a calculator, you find that
tan –1 1 3 1 3
tan –1 (– )
0.322
and –
Therefore, the general solution of the equation is: x or nπ x
– nπ
(where n is any integer)
ANSWER
The specific solutions in the interval 0 ≤ x <2π are: x
x – π
x π
x – π

7. 1. Find the general solution of the equation
2 sin x + 4 = 5.
2 sin x + 4 = 5
Write original equation.
2 sin x = 1
Subtract 4 from each side.
sin x = 1 2
Divide both sides by 2.
sin-1 x = π 3 5π 6 ,
Use a calculator to find both solutions within 0 < x < 2π
ANSWER π 3
+ 2n π or n π 5π 6

8. 2. Solve the equation 3 csc2 x = 4 in the interval 0 ≤ x <2π.
Write original equation.
csc2 x = 4 3
Divide both sides by 3 2 √3
csc x =
Take the square root of both sides =
sin x 1 2 √3
Reciprocal identity 2 √3
sin x =
Standard Form 2 √3
sin-1 x =
Use a calculator to find all solutions within 0 < x < 2π
ANSWER
, , , π 3 2π 4π 5π

9. Oceanography
The water depth d for the Bay of Fundy can be modeled by d π
6.2 t
= 35 – 28 cos
where d is measured in feet and t is the time in hours. If t = 0 represents midnight, at what time(s) is the water depth 7 feet?

10. SOLUTION
Substitute 7 for d in the model and solve for t.
35 – 28 cos π
6.2 t 7 =
Substitute 7 for d.
– 28 cos π
6.2 t
–28 =
Subtract 35 from each side. π
6.2 t
cos 1 =
Divide each side by –28. π
6.2 t =
2nπ
cos q = 1 when q = 2nπ.
t =
12.4n =
Solve for t.
ANSWER
On the interval 0≤ t ≤ 24 (representing one full day), the water depth is 7 feet when t = 12.4(0) = 0 (that is, at midnight) and when t = 12.4(1) = 12.4 (that is, at 12:24 P.M.).

11. 10.4 Assignment
Page 639, 3-21 all

12. 10.4 Solve Trigonometric Equations

13. Write original equation. sin x (sin2 x – 4) = Factor out sin x.
SOLUTION
sin3 x – 4 sin x =
Write original equation.
sin x (sin2 x – 4) =
Factor out sin x.
sin x (sin x + 2)(sin x – 2) =
Factor difference of squares.
Set each factor equal to 0 and solve for x, if possible.
sin x
= 0
sin x + 2
= 0
sin x – 2
= 0 x
= 0 or
= π
sin x
= –2
sin x
= 2
The only solutions in the interval 0 ≤ x ≤ 2π, are x = 0 and x = π.
The general solution is x = 2nπ or x = π + 2nπ where n is any integer.
ANSWER
The correct answer is D.

14. Eliminate solutions
Because sin x is never less than −1 or greater than 1, there are no solutions of sin x = −2 and sin x = 2. The same is true with the cosine of x is never greater than 1.

15. Use the quadratic formula
Solve cos2 x – 5 cos x + 2 = 0 in the interval 0 ≤ x <π.
Because the equation is in the form au2 + bu + c = 0 where u = cos x, you can use the quadratic formula to solve for cos x.
SOLUTION
cos2 x – 5 cos x + 2 =
Write original equation. =
– (–5) + (–5)2 – 4(1)(2)
2(1) –
cos x
Quadratic formula 2 – =
Simplify.
4.56 or 0.44
Use a calculator. x
= cos –
Use inverse cosine.
No solution
1.12
Use a calculator, if possible
ANSWER
In the interval 0 ≤ x ≤ π, the only solution is x

16. Solve an equation with an extraneous solution
Solve 1 + cos x = sin x in the interval 0 ≤ x < 2π.
1 + cos x
sin x =
Write original equation.
(1 + cos x)2
(sin x)2 =
Square both sides.
1 + 2 cos x + cos2 x
sin2 x =
Multiply.
1 + 2 cos x + cos2 x
1– cos2 x =
Pythagorean identity
2 cos2 x + 2 cos x =
Quadratic form
2 cos x (cos x + 1) =
Factor out 2 cos x.
2 cos x
= or
cos x + 1
= 0
Zero product property
cos x
= or
= –1
Solve for cos x.
On the interval 0 ≤ x <2π, cos x = 0 has two solutions: x π 2
= or 3π =

17. On the interval 0 ≤ x <2π, cos x = –1 has one solution: x = π.
Therefore, 1 + cos x = sin x has three possible solutions: x π 2
= ,
π, and 3π
CHECK
To check the solutions, substitute them into the original equation and simplify.
1 + cos x
= sin x
1 + cos x
= sin x
1 + cos x
= sin x
1 + cos π 2
= sin ?
1 + cos π ?
= sin π 3π 2
1 + cos ?
= sin
1 + 0 ?
= 1
1 + (–1) ?
= 0
1 + 0 ?
= –1 1 
= 1 
= 0 1
= –1
ANSWER

18. Graphs of each side of the original equation confirm the solutions.
ANSWER
Graphs of each side of the original equation confirm the solutions.

19. Find the general solution of the equation.
4. sin3 x – sin x = 0
sin3 x – sin x = 0
Write original equation.
sin x(sin2 x – 1) = 0
Factor
sin x(– cos2 x ) = 0
Pythagorean Identity
sin x = 0 OR (– cos2 x ) = 0
Zero product property
sin x = 0 OR cos x = 0
Solve
On the interval 0 ≤ x <2π, sin x = 0 has two solutions: x = 0, π.
On the interval 0 ≤ x <2π, cos x = 0 has two solutions: x
= , π 2 3π
0 + n π or n π
ANSWER π 2

20. Find the general solution of the equation.
– cos x = sin x 3
1 – cos x = sin x 3
Write original equation.
1 – 2 cos x + cos2 x = 3 sin2 x
Square both sides.
1 – 2 cos x + cos2 x = 3 (1 – cos2 x)
Pythagorean Identity
1 – 2 cos x + cos2 x = 3 – 3 cos2 x
Distributive Property
– 2 – 2 cos x + 4 cos2 x = 0
Group like terms
– 2(1 + cos x – 2 cos2 x) = 0
Factor
– 2( cos x) (1 – cos x) = 0
Factor
( cos x) = 0 OR (1 – cos x) = 0
Zero product property
cos x = OR cos x = 1 1 2 –
Solve for cos x

21. 1 2
On the interval 0 ≤ x < 2π, cos x = – has two solutions x = 2π 3 4π 3
On the interval 0 ≤ x < 2π, cos x = 1 has one solution: x
= 0.
But sine is negative in Quadrant III, so is not a solution 4π 3 2π 3
0 + 2n π or n π
ANSWER

22. Solve the equation in the interval 0 ≤ x < π. 6. 2 sin x = csc x
Write original equation.
2 sin x =
sin x 1
Reciprocal identity
2 sin2 x = 1
Multiply both sides by sin x
sin2 x = 2 1
Divide both sides by 2
sin x = 2 √2
Take the square root of both sides π 4 3π 4 ,
ANSWER π 4 3π ,

23. Solve the equation in the interval 0 ≤ x <π.
7. tan2 x – sin x tan2 x = 0
tan2 x – sin x tan2 x = 0
Write original equation.
tan2 x (1 – sin x) = 0
Factor
tan2 x = 0 OR (1 – sin x) = 0
Zero product property
tan x = 0 OR sin x = 1
Simplify
On the interval 0 ≤ x < π, tan x = 0 has two solutions x = 0, π
0, π or
ANSWER π 2

25. 10-4 Assignment day 2
Page 639, all