Evaluate each inverse trigonometric function. Warm Up Solve. 1. x2 + 3x – 4 = 0 2. 3x2 + 7x = 6 Evaluate each inverse trigonometric function. 3. sin-1 1 4. Sin-1 x = 1 or – 4 90° – 60

Objective Solve equations involving trigonometric functions.

Example 1: Solving Trigonometric Equations with Infinitely Many Solutions Find all the solutions of sinθ = sinθ + Method 1 Use algebra. Solve for θ over the principal value of sine, –90° ≤ θ ≤ 90°. sinθ = sinθ + sinθ sinθ = Subtract sinθ from both sides. sinθ = Combine like terms.

Example 1 Continued sinθ = Multiply by 2. θ = sin-1 Apply the inverse sineθ. Find θ when sinθ = θ = 30° Find all real number value of θ, where n is an integer. θ = 30° + 360°n Use the period of the sine function. Use reference angles to find other values of θ. θ = 150° + 360°n

Example 1 Continued Method 2 Use a graph. 1 –1 –90 Graph y = sinθ and y = sinθ + in the same viewing window for –90° ≤ θ ≤ 90°. 90 Use the intersect feature of your graphing calculator to find the points of intersection. The graphs intersect at θ = 30°. Thus, θ = 30° + 1360°n, where n is an integer.

Find all of the solutions of 2cosθ + = 0. Method 1 Use algebra. Solve for θ over the principal value of sine, 0 ≤ θ ≤ . 2cosθ = Subtract from both sides. cosθ = Divide both sides by 2. θ = cos-1 – Apply the inverse cosineθ. Find θ when cosine θ = . θ = 150° θ = 150° + 360°n, 210° +360°n.

Example 2A: Solving Trigonometric Equations in Quadratic Form Solve each equation for the given domain. 4tan2θ – 7 tanθ + 3 = 0 for 0° ≤ θ ≤ 360°. 4tan2θ – 7 tanθ + 3 = 0 Factor the quadratic expression by comparing it with 4x2 – 7x + 3 = 0. (tanθ – 1)(4tanθ – 3) = 0 Apply the Zero Product Property.

Example 2A Continued tanθ = 1 or tan θ = Apply the inverse tangent. θ = tan-1(1) θ = tan-1 Use a calculator. Find all angles for 0°≤ θ ≤360°. = 45° or 225° ≈ 36.9° or 216.9°

Example 2B: Solving Trigonometric Equations in Quadratic Form 2cos2θ – cosθ = 1 for 0 ≤ θ ≤ . 2cos2θ – cosθ – 1 = 0 Subtract 1 from both sides. Factor the quadratic expression by comparing it with 2x2 – x + 1 = 0. (2cosθ + 1) (cosθ – 1) = 0 cosθ = or cosθ = 1 Apply the Zero Product Property. θ = or θ = 0 Find both angles for 0 ≤ θ ≤ .

Solve each equation for 0 ≤ θ ≤ 2. cos2 θ + 2cosθ = 3 Subtract 3 from both sides. cos2 θ + 2cosθ – 3 = 0 Factor the quadratic expression by comparing it to x2 +2x – 3 = 0. (cosθ – 1)(cosθ + 3) = 0 Apply the Zero Product Property. cosθ = 1 or cosθ = –3 cosθ = – 3 has no solution because –3 ≤ cosθ ≤ 1. The only solution will come from cosθ = 1. cosθ = 2 or 0

Solve each equation for 0 ≤ θ ≤ 2. Example 2b Solve each equation for 0 ≤ θ ≤ 2. sin2θ + 5 sinθ – 2 = 0 The equation is in quadratic form but can not be easily factored. Use the quadratic formula. sinθ = Apply the inverse sine. Use a calculator. Find both angles.

Example 5 – Solving Trigonometric Equations Find all solutions of the equation. (a) 2 sin  – 1 = 0 (b) tan2  – 3 = 0 Solution: (a) We start by isolating sin  : 2 sin  – 1 = 0 2 sin  = 1 sin  = Given equation Add 1 Divide by 2 The solutions are  = + 2k  = + 2k where k is any integer.

(b) We start by isolating tan  : tan2  – 3 = 0 tan2  = 3 cont’d Because tangent has period , we first find the solutions in any interval of length . In the interval (– /2,  /2) the solutions are  =  /3 and  = – /3. To get all solutions, we add integer multiples of  to these solutions:  = + k  = – + k where k is any integer.

Solve the equation 2 cos2  – 7 cos  + 3 = 0. Solution: We factor the left-hand side of the equation. 2 cos2  – 7 cos  + 3 = 0 (2 cos  – 1)(cos  – 3) = 0 2 cos  – 1 = 0 or cos  – 3 = 0 cos  = or cos  = 3 Given equation Factor Set each factor equal to 0 Solve for cos 

Because cosine has period 2, we first find the solutions in the interval [0, 2). For the first equation the solutions are  =  /3 and  = 5 /3 (see Figure 7). Figure 7

The second equation has no solution because cos  is never greater than 1. Thus the solutions are  = + 2k  = + 2k where k is any integer.

Example 7 – Solving a Trigonometric Equation by Factoring Solve the equation 5 sin  cos  + 4 cos  = 0. Solution: We factor the left-hand side of the equation: 5 sin  cos  + 2 cos  = 0 cos  (5 sin  + 2) = 0 cos  = 0 or 5 sin  + 4 = 0 sin  = –0.8 Given equation Factor Set each factor equal to 0 Solve for sin 

Example 7 – Solution cont’d Because sine and cosine have period 2, we first find the solutions of these equations in an interval of length 2. For the first equation the solutions in the interval [0, 2) are  =  /2 and  = 3 /2 . To solve the second equation, we take sin–1 of each side: sin  = –0.80  = sin–1(–0.80)

Example 7 – Solution cont’d   –0.93 So the solutions in an interval of length 2 are  = –0.93 and  =  + 0.93  4.07 (see Figure 8). Calculator (in radian mode) Figure 8

Example 7 – Solution cont’d We get all the solutions of the equation by adding integer multiples of 2 to these solutions.  = + 2k,  = + 2k,   –0.93 + 2k,   4.07 + 2k where k is any integer.

HW 28 page 396 #1, 5-15 odd, 21-25 odd, 26, 29, 30