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EXAMPLE 1

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. =. =. 2 π. –. = π. =. 3. EXAMPLE 1. Solve a trigonometric equation. SOLUTION.

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EXAMPLE 1

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  1. 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 EXAMPLE 1 Solve a trigonometric 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

  2. 2π + 2nπ = = π 3 + 2nπ 3 You can check the answer by graphing y = sinxand y = in the same coordinate plane. Then find the points where the graphs intersect. You can see that there are infinitely many such points. 3 2 EXAMPLE 1 Solve a trigonometric equation 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

  3. EXAMPLE 1 Solve a trigonometric equation

  4. = 3 = 1 = 1 1 1 1 + = – 3 3 3 9 tan –1 0.322 and tan–1 (– ) – 0.322. 0.322 + nπ or – 0.322 + nπ EXAMPLE 2 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. tanx Take square roots of each side. Using a calculator, you find that Therefore, the general solution of the equation is: x x (where nis any integer)

  5. 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 EXAMPLE 2 Solve a trigonometric equation in an interval

  6. Oceanography The water depth dfor the Bay of Fundy can be modeled by d = 35 – 28 cos π t 6.2 EXAMPLE 3 Solve a real-life trigonometric equation 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?

  7. 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 EXAMPLE 3 Solve a real-life trigonometric equation 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.

  8. 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.). EXAMPLE 3 Solve a real-life trigonometric equation

  9. ANSWER 4π 2π 5π 5π 3 3 6 3 + 2n πor + 2n π π π 3 3 , , , ANSWER for Examples 1, 2, and 3 GUIDED PRACTICE 1. Find the general solution of the equation 2sinx + 4 = 5. 2. Solve the equation 3csc2x = 4 in the interval 0 ≤ x <2π. 3. OCEANOGRAPHY: In Example 3, at what time(s) is the water depth 63 feet? 6:12 A.M and6:36P.M. ANSWER

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