# EXAMPLE 1 Solve a trigonometric equation Solve 2 sin x – 3 = 0. SOLUTION First isolate sin x on one side of the equation. Write original equation. 2 sin.

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EXAMPLE 1 Solve a trigonometric equation Solve 2 sin x – 3 = 0. SOLUTION First isolate sin x on one side of the equation. Write original equation. 2 sin x – 3 0 = Add 3 to each side. 2 sin x3= Divide each side by 2. sin x3 2 = One solution of sin x = 3 2 in the interval 0 ≤ x < 2 π is x 3 2 = sin –1 π 3. = The other solution in the interval is x π 3 – = π 2π2π 3. =

EXAMPLE 1 Solve a trigonometric equation 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: x + 2nπ π 3 = or x + 2nπ 2π2π 3 = (where n is any integer) CHECKYou can check the answer by graphing y = sin x and 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

EXAMPLE 2 Solve a trigonometric equation in an interval Solve 9 tan 2 x + 2 = 3 in the interval 0 ≤ x <2π. Write original equation. 9 tan 2 x + 23= Subtract 2 from each side. 9 tan 2 x1= Divide each side by 9. tan 2 x 1 9 = Take square roots of each side. tan x 1 3 + – = Using a calculator, you find that tan –1 1 3 0.322 and 1 3 tan –1 (– ) – 0.322. Therefore, the general solution of the equation is: x or 0.322 + nπx– 0.322 + nπ (where n is any integer)

EXAMPLE 2 Solve a trigonometric equation in an interval ANSWER The specific solutions in the interval 0 ≤ x <2π are: x 0.322x – 0.322 + π 2.820 x 0.322 + π 3.464x – 0.322 + 2π 5.961

EXAMPLE 3 Solve a real-life trigonometric equation 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?

EXAMPLE 3 Solve a real-life trigonometric equation SOLUTION Substitute 7 for d in the model and solve for t. Substitute 7 for d. 35 – 28 cos π 6.2 t 7= Subtract 35 from each side. –28 cos π 6.2 t –28 = Divide each side by –28. π 6.2 t cos 1 = cos  = 1 when  = 2nπ. π 6.2 t = 2nπ Solve for t. t =12.4n=

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

GUIDED PRACTICE for Examples 1, 2, and 3 1. Find the general solution of the equation 2sin x + 4 = 5. ANSWER π 3 + 2n π or + 2n π 5π5π 6 2. Solve the equation 3csc 2 x = 4 in the interval 0 ≤ x <2π. ANSWER,,, π 3 2π2π 3 4π4π 3 5π5π 3 3. OCEANOGRAPHY: In Example 3, at what time(s) is the water depth 63 feet? ANSWER 6:12 A.M and 6:36 P.M.

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