1. Chapter 6
Equations 6.1 Solving Trigonometric Equations 6.2 More on Trigonometric Equations 6.3 Trigonometric Equations Involving Multiples Angles 6.4 Parametric Equations and Further Graphing

2. 6.1 Solving Trigonometric Equations
Decide whether the equation is linear or quadratic in form, so you can determine the solution method.
If only one trigonometric function is present, first solve the equation for that function.
If more than one trigonometric function is present, rearrange the equation so that one side equals 0. Then try to factor and set each factor equal to 0 to solve.

3. If the equation is quadratic in form, but not factorable, use the quadratic formula. Check that solutions are in the desired interval.
Try using identities to change the form of the equation. If may be helpful to square both sides of the equation first. If this is done, check for extraneous solutions.

4. Example: Linear Method
Solve 2 cos2 x  1 = 0
Solution: First, solve for cos x on the unit circle.

5. Example: Factoring Solve 2 cos x + sec x = 0 Solution
Since neither factor of the equation can equal zero, the equation has no solution.

6. Example: Factoring continued
Solve 2 sin2 x + 3sin x + 1 = 0 (2 Sin x + 1)(Sin x +1) = 0
Set each factor equal to 0. a) 2 Sin x + 1 = 0 b) Sin x + 1 = 0
continued

7. Example: Factoring continued

8. Example: Squaring Solve cos x + 1 = sin x [0, 2]
Check the solutions in the original equation. The only solutions are /2 and .

9. a) over the interval [0, 2), and b) give all solutions Solution:
6.2 More on Trigonometric Equations
a) over the interval [0, 2), and
b) give all solutions
Solution:
Write the interval as the inequality

10. Example: Using a Half-Angle Identity continued
The corresponding interval for x/2 is
Solve
Sine values that corresponds to 1/2 are

11. Example: Using a Half-Angle Identity continued
b) Sine function with a period of 4, all solutions are given by the expressions
where n is any integer.

12. Example: Double Angle Solve cos 2x = cos x over the interval [0, 2).
First, change cos 2x to a trigonometric function of x. Use the identity

13. Example: Double Angle continued
Over the interval

14. Example: Multiple-Angle Identity
Solve over the interval [0, 360).

15. List all solutions in the interval.
6.3 Trigonometric Equations Involving Multiples Angles
List all solutions in the interval.
The final two solutions were found by adding 360 to 60 and 120, respectively, giving the solution set

16. Example: Multiple Angle
Solve tan 3x + sec 3x = 2 over the interval [0, 2).
Tangent and secant are related so use the identity

17. Example: Multiple Angle continued

18. Example: Multiple Angle continued
Use a calculator and the fact that cosine is positive in quadrants I and IV,
Since both sides of the equation were squared, each proposed solution must be checked. The solution set is {.2145, , }.

19. Solving for x in Terms of y Using Inverse Function
Example: y = 3 cos 2x for x.
Solution: We want 2x alone on one side of the equation so we can solve for 2x, and then for x.

20. Solving an Equation Involving an Inverse Trigonometric Function
Example: Solve 2 arcsin
Solution: First solve for arcsin x, and then for x.
The solution set is {1}.

21. Solving an Equation Involving Inverse Trigonometric Functions
Example: Solve
Solution: Let Then sin and for u in
quadrant I, the equation becomes

22. Solving an Equation Involving Inverse Trigonometric Functions continued
Sketch a triangle and label it using the facts that u is in quadrant I and
Since x = cos u, x = and the solution set is { }.

23. Solving an Inverse Trigonometric Equation Using an Identity
Example: Solve
Solution: Isolate one inverse function on one side of the equation.
(1)

24. Solving an Inverse Trigonometric Equation Using an Identity continued
Let u = arccos x, so 0  u  by definition.
(2)
Substitute this result into equation (2) to get
(3)

25. Solving an Inverse Trigonometric Equation Using an Identity continued
From equation (1) and by the definition of the arcsine function,
Since we must have
Thus x > 0. From this triangle we find that

26. Solving an Inverse Trigonometric Equation Using an Identity continued
Now substituting into equation (3) using
The solution set is { }.

27. 6.4 Parametric Equations and Further Graphing
A plane curve is a set of points (x, y) such that x = f(t), y = g(t), and f and g are both defined on an interval I. The equations x = f(t) and y = g(t) are parametric equations with parameter t.

28. Graphing a Plane Curve Defined Parametrically
Example: Let x = t2 and y = 2t + 3, for t in [3,3]. Graph the set of ordered pairs (x, y).
Solution: Make a table of
corresponding values of
t, x, and y over the domain of t. 9 3 7 4 2 5 1 1 2 3 y x t

29. Graphing a Plane Curve Defined Parametrically continued
Plotting the points shows a graph of a portion of a parabola with horizontal axis y = 3. The arrowheads indicate the direction the curve traces as t increases.

30. Finding an Equivalent Rectangular Equation
Example: Find a rectangular equation for the plane curve of the previous example defined as follows x = t2 , y = 2t + 3, for t in [3, 3]
Solution: Solve either equation for t.

31. Finding an Equivalent Rectangular Equation continued
Now substitute this result into the first equation to get
This is the equation of a horizontal parabola opening to the right. Because t is in [3, 3], x is in [0, 9] and y is in [3, 9]. This rectangular equation must be given with its restricted domain as 4x = (y  3)2 , for x in [0, 9].

32. Graphing a Plane Curve Defined Parametrically
Example: Graph the plane curve defined by x = 2 sin t,
y = 3cos t, for t in [0,2 ].
Solution: Use the fact that sin2 t + cos2t = 1. Square both
sides of each equation; solve one for sin2 t, the
other for cos2t.

33. Graphing a Plane Curve Defined Parametrically continued
Now add corresponding sides of the two equations.
This is the equation of an ellipse.

34. Finding Alternative Parametric Equation Forms
Give two parametric representations for the equation of the parabola y = (x + 5)2 +3.
Solution:
The simplest choice is to let x = t, y = (t + 5) for t in (, )
Another choice, which leads to a simpler equation for y, is
x = t + 5, y = t for t in (, ).

35. Application
A small rocket is launched from a table that is 3.36 ft above the ground. Its initial velocity is 64 ft per sec, and it is launched at an angle of 30° with respect to the ground. Find the rectangular equation that models its path. What type of path does the rocket follow?
Solution: The path of the rocket is defined by the parametric equations
x = (64 cos 30°)t and y = (64 sin 30°)t  16t
Or equivalently,

36. Application continued
From we obtain
Substituting into the other parametric equations for t yields
Simplifying, we find that the rectangular equation is
Because the equation defines a parabola, the rocket follows a parabolic path.