10.4 Solve Trigonometric Equations  

Solving trig equations… What’s the difference?  

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

(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  

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 – 0.322. Therefore, the general solution of the equation is: x or 0.322 + nπ x – 0.322 + nπ (where n is any integer) ANSWER The specific solutions in the interval 0 ≤ x <2π are: x 0.322 x – 0.322 + π 2.820 x 0.322 + π 3.464 x – 0.322 + 2π 5.961

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 + 2n π 5π 6

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

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?

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.).

10.4 Assignment Page 639, 3-21 all

10.4 Solve Trigonometric Equations  

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.

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.

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 5 + 17 2 – = Simplify. 4.56 or 0.44 Use a calculator. x = cos –1 0.44 Use inverse cosine. No solution 1.12 Use a calculator, if possible ANSWER In the interval 0 ≤ x ≤ π, the only solution is x 1.12.

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 = 0 or cos x + 1 = 0 Zero product property cos x = 0 or = –1 Solve for cos x. On the interval 0 ≤ x <2π, cos x = 0 has two solutions: x π 2 = or 3π =

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  

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

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

Find the general solution of the equation. 5. 1 – 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(1 + 2 cos x) (1 – cos x) = 0 Factor (1 + 2 cos x) = 0 OR (1 – cos x) = 0 Zero product property cos x = OR cos x = 1 1 2 – Solve for cos x

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 + 2n π ANSWER

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 ,

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

 

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