Solve . Original equation Solve by Isolating Trigonometric Expressions Solve . Original equation Subtract 3cos x from each side to isolate the trigonometric expression. Solve for cos x. Example 1

Solve by Isolating Trigonometric Expressions The period of cosine is 2π, so you only need to find solutions on the interval [0, 2π). The solutions on this interval are . The solutions on the interval (–∞, ∞) are then found by adding integer multiples 2π. Therefore, the general form of the solutions is x = , where n is an integer. Answer: Example 1

Solve sin x + = – sin x. A. B. C. D. Example 1

3 tan2 x – 4 = –3 Original equation 3 tan2 x = 1 Add 4 to each side. Solve by Taking the Square Root of Each Side Solve 3 tan2 x – 4 = –3. 3 tan2 x – 4 = –3 Original equation 3 tan2 x = 1 Add 4 to each side. Divide each side by 3. Take the square root of each side. Rationalize the denominator. Example 2

Solve by Taking the Square Root of Each Side The period of tangent is π. On the interval [0, π), tan x = when x = and tan x = when x = . The solutions on the interval (–∞, ∞) have the general form , where n is an integer. Answer: Example 2

Solve 5 tan2x – 15 = 0. A. B. C. D. Example 2

A. Find all solutions of on the interval [0, 2π). Solve by Factoring A. Find all solutions of on the interval [0, 2π). Original equation Isolate the trigonometric terms. Factor. Zero Product Property Example 3

On the interval [0, 2π), the equation has solutions . Solve by Factoring Solve for cos x. Solve for x on [0, 2π). On the interval [0, 2π), the equation has solutions . Answer: Example 3

Solve by Factoring B. Find all solutions of 2sin2x + sinx – 1 = 0 on the interval [0, 2π). Original equation Factor. Zero Product Property Solve for sin x. Solve for x on [0, 2π). Example 3

Solve by Factoring On the interval [0, 2π), the equation 2sin2x + sinx – 1 = 0 has solutions . Answer: Example 3

Find all solutions of 2 tan4 x – tan2 x – 15 = 0 on the interval [0, π). B. C. D. Example 3

Trigonometric Functions of Multiple Angles PROJECTILES A projectile is sent off with an initial speed vo of 350 m/s and clears a fence 3000 m away. The height of the fence is the same height as the initial height of the projectile. If the distance the projectile traveled is given by , find the interval of possible launch angles to clear the fence. Example 4

Definition of inverse sine. Trigonometric Functions of Multiple Angles Original formula d = 3000 and v0 = 350 Simplify. Multiply each side by 9.8. Divide each side by 122,500. Definition of inverse sine. Example 4

sin–10.24 = 2 Definition of inverse sine Trigonometric Functions of Multiple Angles Recall from Lesson 4-6 that the range of the inverse sine function is restricted to acute angles of θ in the interval [–90°, 90°]. Since we are finding the inverse sine of 2θ instead of θ, we need to consider angles in the interval [–2(90°), 2(90°)] or [–180°, 180°]. Use your calculator to find the acute angle and the reference angle relationship sin (180° − θ) = sin θ to find the obtuse angle. sin–10.24 = 2 Definition of inverse sine 13.9° or 166.1° = 2 sin–1(0.24) ≈13.9° and sin(180° – 13.9°) = 166.1° 7.0° or 83.1° =  Divide by 2. Example 4

Trigonometric Functions of Multiple Angles The interval is [7.0°, 83.1°]. The ball will clear the fence if the angle is between 7.0° and 83.1°. Answer: 7.0° ≤  ≤ 83.1° CHECK Substitute the angle measures into the original equation to confirm the solution. Original formula  = 7.0° or  = 83.1°  Use a calculator. Example 4

GOLF A golf ball is sent off with an initial speed vo of 36 m/s and clears a small barricade 70 m away. The height of the barricade is the same height as the initial height of the ball. If the distance the ball traveled is given by , find the interval of possible launch angles to clear the barricade. A. 1.6° ≤  ≤ 88.5° B. 3.1° ≤  ≤ 176.9° C. 16.0° ≤  ≤ 74.0° D. 32° ≤  ≤ 148.0° Example 4

sin2 x – sin x + 1 = cos2 x Original equation Solve by Rewriting Using a Single Trigonometric Function Find all solutions of sin2 x – sin x + 1 = cos2 x on the interval [0, 2π). sin2 x – sin x + 1 = cos2 x Original equation –cos2 x + sin2 x – sin x + 1 = 0 Subtract cos2 x from each side. –(1 – sin2 x) + sin2 x – sin x + 1 = 0 Pythagorean Identity 2sin2 x – sin x = 0 Simplify. sin x (2sin x – 1) = 0 Factor. Example 5

sin x = 0 2sin x – 1 = 0 Zero Product Property Solve by Rewriting Using a Single Trigonometric Function sin x = 0 2sin x – 1 = 0 Zero Product Property 2sin x = 1 Solve for sin x. Solve for x on [0, 2π). x = 0, π Example 5

Solve by Rewriting Using a Single Trigonometric Function Answer: CHECK The graphs of Y1 = sin2 x – sin x + 1 and Y2 = cos2 x intersect at on the interval [0, 2π) as shown.  Example 5

Find all solutions of 2sin2x = cosx + 1 on the interval [0, 2). B. C. D. Example 5

Find all solutions of sin x – cos x = 1 on the interval [0, 2π). Solve by Squaring Find all solutions of sin x – cos x = 1 on the interval [0, 2π). sin x – cos x = 1 Original equation sin x = cos x + 1 Add cos x to each side. sin2 x = cos2 x + 2cos x + 1 Square each side. 1 – cos2 x = cos2 x + 2cos x + 1 Pythagorean Identity 0 = 2cos2 x + 2cos x Subtract 1 – cos2x from each side. 0 = cos2 x + cos x Divide each side by 2. 0 = cos x(cos x + 1) Factor. Example 6

cos x = 0 cos x + 1 = 0 Zero Product Property Solve by Squaring cos x = 0 cos x + 1 = 0 Zero Product Property cos x = –1 Solve for cos x. , x = π Solve for x on [0, 2). Original formula Substitute sin π – cos π = 1  Simplify. Example 6

Therefore, the only valid solutions are on the interval . Solve by Squaring Therefore, the only valid solutions are on the interval . Answer: Example 6

Find all solutions of 1 + cos x = sin x on the interval [0, 2π). B. C. D. Example 6

End of the Lesson