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

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.

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.

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

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

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

Example: Factoring continued

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

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

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

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.

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

Example: Double Angle continued Over the interval

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

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

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

Example: Multiple Angle continued

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, 2.3089, 4.4033}.

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.

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}.

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

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 { }.

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

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)

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

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

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.

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

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.

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.

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].

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.

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

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)2 + 3 for t in (, ) Another choice, which leads to a simpler equation for y, is x = t + 5, y = t2 + 3 for t in (, ).

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  16t2 + 3.36 Or equivalently,

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.