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10.4 Solve Trigonometric Equations

10.4 Solve Trigonometric Equations. Solving trig equations… What’s the difference?. Solve 2 sin x – 3 = 0. 0. =. π. π. =. 3. Add 3 to each side. 3. 3. 3. 3. 3. =. 2. 2. 2. 2 sin x – 3. One solution of sin x =. in the interval 0 ≤ x < 2 π is. . sin –1. =.

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10.4 Solve Trigonometric Equations

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  1. 10.4 Solve Trigonometric Equations

  2. Solving trig equations…What’s the difference?

  3. Solve 2 sinx – 3 = 0. 0 = π π = 3 Add 3 to each side. 3 3 3 3 3 = 2 2 2 2 sinx – 3 One solution of sin x = in the interval 0 ≤ x <2 πis . sin–1 = = 2π – = π . = 3 Solve a trig equation SOLUTION First isolate sin xon one side of the equation. Write original equation. 2 sin x sin x Divide each side by 2. x The other solution in the interval is x

  4. 2π + 2nπ = = π 3 + 2nπ 3 Moreover, because y =sin xis periodic, there will be infinitely many solutions. You can use the two solutions found above to write the general solution: (where nis any integer) x x or CHECK

  5. = 3 The specific solutions in the interval 0 ≤ x <2π are: ANSWER = 1 x – 0.322 + 2π 5.961 x – 0.322 + π 2.820 x 0.322 x 0.322 + π 3.464 = 1 1 1 1 + = – 3 3 9 3 tan –1 0.322 and tan–1 (– ) – 0.322. 0.322 + nπ or – 0.322 + nπ Solve a trigonometric equation in an interval Solve 9 tan2x + 2 = 3 in the interval 0 ≤ x <2π. 9 tan2x + 2 Write original equation. 9 tan2x Subtract 2 from each side. tan2x Divide each side by 9. Take square roots of each side. tanx Using a calculator, you find that Therefore, the general solution of the equation is: x x (where nis any integer)

  6. ANSWER 5π 6 + 2n πor + 2n π sinx = π 1 5π π 3 2 6 3 , sin-1x = 1. Find the general solution of the equation 2 sinx + 4 = 5. 2 sinx + 4 = 5 Write original equation. Subtract 4 from each side. 2 sinx = 1 Divide both sides by 2. Use a calculator to find both solutions within 0 < x < 2π

  7. 2π 4π 5π csc2x = 3 3 3 1 csc x = 2 2 π sin x 4 3 3 = √3 √3 √3 √3 2 2 sin x = sin-1x = , , , ANSWER 2. Solve the equation 3 csc2x = 4 in the interval 0 ≤ x <2π. 3 csc2x = 4 Write original equation. Divide both sides by 3 Take the square root of both sides Reciprocal identity Standard Form Use a calculator to find all solutions within 0 < x < 2π

  8. Oceanography The water depth dfor the Bay of Fundy can be modeled by d = 35 – 28 cos π t 6.2 where dis measured in feet and tis the time in hours. If t= 0 represents midnight, at what time(s) is the water depth 7 feet?

  9. ANSWER On the interval 0≤t ≤ 24 (representing one full day), the water depth is 7feet when t = 12.4(0) = 0 (that is, at midnight) and when t = 12.4(1) = 12.4 (that is, at 12:24P.M.). 35 – 28 cos = 7 –28 – 28 cos = π π π π t t t t 6.2 6.2 6.2 6.2 1 cos = 2nπ = t = = 12.4n SOLUTION Substitute 7 for din the model and solve for t. Substitute 7 for d. Subtract 35 from each side. Divide each side by –28. cos q = 1whenq = 2nπ. Solve for t.

  10. 10.4 Assignment Page 639, 3-21 all

  11. 10.4 Solve Trigonometric Equations

  12. 0 sin3x – 4 sinx = ANSWER The correct answer is D. 0 sinx (sin2x –4) = sin x (sinx + 2)(sinx – 2) 0 = sinx + 2 sinx – 2 sinx = 0 = 0 = 0 x sinx sinx = 0 or x = π = –2 = 2 SOLUTION Write original equation. Factor out sin x. Factor difference of squares. Set each factor equal to 0 and solve for x, if possible. The only solutions in the interval 0 ≤ x ≤2π, are x = 0 and x = π. The general solution is x = 2nπor x = π + 2nπ where nis any integer.

  13. 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.

  14. x = cos –1 0.44 1.12 In the interval 0 ≤ x ≤ π, the only solution isx 1.12. ANSWER 0 = – = – (–5) + (–5)2 – 4(1)(2) 5 + 17 – 2(1) = 2 4.56 or 0.44 Use the quadratic formula Solvecos2x – 5 cosx + 2 = 0 in the interval 0 ≤ x <π. Because the equation is in the form au2 + bu + c = 0 where u = cosx, you can use the quadratic formula to solve for cos x. SOLUTION cos2x –5 cosx + 2 Write original equation. cos x Quadratic formula Simplify. Use a calculator. Use inverse cosine. No solution Use a calculator, if possible

  15. 1 + cosx sinx = (1 + cosx)2 (sin x)2 = 1 + 2 cosx + cos2x = sin2x 1 + 2 cosx + cos2x = 1– cos2x π 2 cos2x + 2 cosx = 0 2 2 cosx (cosx + 1) 0 = 2 cos x cosx + 1 = 0 or = 0 cos x = 0 or cosx = –1 3π x = x = or 2 Solve an equation with an extraneous solution Solve 1 + cosx = sinxin the interval 0 ≤ x < 2π. Write original equation. Square both sides. Multiply. Pythagorean identity Quadratic form Factor out 2cosx. Zero product property Solve for cos x. On the interval 0 ≤ x <2π, cosx = 0 has two solutions:

  16. ANSWER x = π. Therefore, 1 + cosx = sinxhas three possible solutions: π 2 x = , π, and π π 2 2 1 + cosx 1 + cosx 1 + cos x = sin x = sin x = sin x ? ? 3π ? 3π 1 + cos π 1 + cos 1 + cos = sin π = sin = sin 2 2 ? ? ? 1 + 0 =1 1 + (–1) = 0 1 + 0 =–1 0 = 0 1  = 1 1 = –1  3π 2 On the interval 0 ≤ x <2π, cosx = –1 has one solution: CHECK To check the solutions, substitute them into the original equation and simplify.

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

  18. x = , 3π π π 2 2 2 On the interval 0 ≤ x <2π, cosx = 0 has two solutions: ANSWER 0 + n πor + n π Find the general solution of the equation. 4.sin3x – sinx = 0 sin3x – sinx = 0 Write original equation. sin x(sin2x – 1) = 0 Factor sin x(– cos2x ) = 0 Pythagorean Identity sin x = 0OR (– cos2x ) = 0 Zero product property sin x = 0OR cosx = 0 Solve On the interval 0 ≤ x <2π, sinx = 0 has two solutions:x = 0, π.

  19. 3 3 1 – cosx = sinx 1 – 2 5. 1 – cosx = sinx cosx = OR cosx = 1 Find the general solution of the equation. Write original equation. 1 – 2 cosx + cos2x = 3 sin2x Square both sides. 1 – 2 cosx + cos2x = 3 (1 – cos2 x) Pythagorean Identity 1 – 2 cosx + cos2x = 3 – 3 cos2 x Distributive Property –2 – 2 cosx + 4 cos2x = 0 Group like terms –2(1 + cosx – 2 cos2x) = 0 Factor –2(1 + 2 cosx) (1– cosx)= 0 Factor (1 + 2 cosx) = 0 OR (1– cosx)= 0 Zero product property Solve for cosx

  20. x = 0. On the interval 0 ≤ x < 2π, cosx = 1 has one solution: 2π 3 2π 4π 1 4π But sine is negative in Quadrant III, so is not a solution 3 2 3 3 ANSWER 0 + 2n πor + 2n π On the interval 0 ≤ x < 2π, cosx = – has two solutionsx =

  21. 3π 3π 4 4 1 2 sinx = sinx √2 1 π π 2 2 4 4 sin2x = sinx = , ANSWER Solve the equation in the interval 0 ≤ x < π. 6. 2 sinx = cscx 2 sinx = cscx Write original equation. Reciprocal identity 2 sin2x = 1 Multiply both sides by sin x Divide both sides by 2 Take the square root of both sides ,

  22. π 2 ANSWER 0, πor Solve the equation in the interval 0 ≤ x <π. 7.tan2x – sinx tan2x = 0 tan2x – sinx tan2x = 0 Write original equation. tan2x (1 – sinx) = 0 Factor tan2x = 0OR (1 – sinx) = 0 Zero product property tanx = 0OR sinx = 1 Simplify On the interval 0 ≤ x < π, tanx = 0 has two solutions x = 0, π

  23. 10-4 Assignment day 2 Page 639, 24-35 all

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