This is solving a trigonometric equation for temperature in a daily cycle context. To find when T(t) = 15°C, we isolate the cosine: 10cos(πt/12) + 20 = 15, which gives cos(πt/12) = -0.5. Using inverse cosine, we get πt/12 = 2π/3 radians (120°), so t = 8 hours. Since cosine is negative in both the second and third quadrants, we also have πt/12 = 4π/3 radians (240°), giving t = 16 hours. Option C incorrectly suggests t = 4 and t = 20, which would give cos(π/3) = 0.5 and cos(5π/3) = 0.5, not the required -0.5. When solving cosine equations, remember that cos(θ) = -0.5 occurs at θ = 2π/3 and θ = 4π/3 in one period.

This problem asks us to solve a trigonometric equation for beacon brightness. Setting B(t) = 15 gives 5cos(2t) + 10 = 15, so cos(2t) = 1. The cosine function equals 1 when 2t = 0 or 2t = 2π within one period [0, 2π]. Solving for t gives t = 0 and t = π, which matches answer B. Answer A incorrectly suggests t = π/2, but cos(π) = -1, not 1, so B(π/2) = 5(-1) + 10 = 5, not 15. When cosine equals 1, the angle must be a multiple of 2π. Always check that your solutions fall within the specified interval.

This question tests understanding of how to solve trigonometric equations that arise from modeling real-world contexts and interpret the solutions. To solve a trigonometric equation from a model, first isolate the trigonometric function (sin, cos, or tan) on one side, then use the appropriate inverse trigonometric function (arcsin, arccos, or arctan) to find the reference angle, and finally determine all solutions in the specified domain by considering the periodicity and symmetry of the trig function. Starting with the model 3 cos(π t /4) +5 =6.5, we rearrange to cos(π t /4) =0.5, using the inverse, π t /4 = arccos(0.5) = π/3, so t = (4/π) arccos(1/2) s, which is the first time after t=0. Choice A is correct because it properly isolates the trig function, applies the correct inverse function, and finds the first solution in the domain. Choice B fails by using arcsin(1/2) when the isolated equation has cos, not sin, using the wrong inverse function. Key to solving trig equations in context: (1) isolate the trig function, (2) use the inverse to find the reference angle, (3) find all solutions in the specified interval using periodicity and symmetry, (4) check that solutions make sense in the real-world context. For real-world problems, always interpret your solution in context: t = (4/π) arccos(1/2) seconds isn't just an expression, it's the specific time when the buoy first reaches 6.5 m.

This question tests understanding of how to solve trigonometric equations that arise from modeling real-world contexts and interpret the solutions. When a real-world phenomenon is modeled by a trigonometric function, solving an equation like T(t) = [target] gives the time(s) when the phenomenon reaches that target value, with multiple solutions corresponding to the periodic nature of the phenomenon (it repeats). Starting with the model 10 cos(π t /12) +20 =15, we rearrange to cos(π t /12) = -0.5, the reference angle is arccos(0.5) = π/3, because cosine is negative in quadrants II and III, the solutions in [0, 2π] are 2π/3 and 4π/3, so t =8 h and t=16 h in [0,24). Choice C is correct because it properly isolates the trig function, applies the correct inverse function, finds all solutions in the domain, and interprets correctly in context. Choice A fails by giving times outside the interval or incorrect solutions, misapplying the quadrant adjustments for negative cosine. Key to solving trig equations in context: (1) isolate the trig function, (2) use the inverse to find the reference angle, (3) find all solutions in the specified interval using periodicity and symmetry, (4) check that solutions make sense in the real-world context. Remember that sine and cosine equations typically have two solutions in a 2π interval (one in each of two quadrants where the function has the same value), while tangent equations have one solution per π interval due to its shorter period.

This question tests understanding of how to solve trigonometric equations that arise from modeling real-world contexts and interpret the solutions. To solve a trigonometric equation from a model, first isolate the trigonometric function (sin, cos, or tan) on one side, then use the appropriate inverse trigonometric function (arcsin, arccos, or arctan) to find the reference angle, and finally determine all solutions in the specified domain by considering the periodicity and symmetry of the trig function. Starting with the model 4 cos(3t) =2, we rearrange to cos(3t) =0.5, using the inverse, 3t = arccos(0.5) = π/3 rad, so t = π/9 s, which is the first positive time. Choice B is correct because it properly isolates the trig function, applies the correct inverse function, and finds the first solution in the domain. Choice A fails by using π/6, which would be for arcsin(0.5) instead of arccos(0.5), confusing the inverse functions. Key to solving trig equations in context: (1) isolate the trig function, (2) use the inverse to find the reference angle, (3) find all solutions in the specified interval using periodicity and symmetry, (4) check that solutions make sense in the real-world context. For real-world problems, always interpret your solution in context: t = π/9 seconds isn't just a number, it's the specific time when the mass reaches 2 cm after t=0.

This question tests understanding of how to solve trigonometric equations that arise from modeling real-world contexts and interpret the solutions. To solve a trigonometric equation from a model, first isolate the trigonometric function (sin, cos, or tan) on one side, then use the appropriate inverse trigonometric function (arcsin, arccos, or arctan) to find the reference angle, and finally determine all solutions in the specified domain by considering the periodicity and symmetry of the trig function. The equation that correctly represents finding times when V=5 is 10 sin(120π t)=5, as it directly sets the model equal to the target value. Choice A is correct because it properly sets up the equation by equating the model to 5 without unnecessary additions or incorrect constants. Choice D fails by setting sin(120π t)=10.5, which is impossible since sine cannot exceed 1, likely from misdividing 5/10=0.5 but writing 10.5. Key to solving trig equations in context: (1) isolate the trig function, (2) use the inverse to find the reference angle, (3) find all solutions in the specified interval using periodicity and symmetry, (4) check that solutions make sense in the real-world context. Domain restrictions matter: if the context specifies 'first 12 hours' or 't ≥ 0', eliminate any mathematically valid solutions that fall outside these constraints—negative times usually don't make physical sense.

This problem asks us to solve a trigonometric equation for beacon brightness over a full period. Setting B(t) = 12.5, we get 5sin(2t) + 10 = 12.5, so sin(2t) = 0.5. In the interval [0,2π], we need all solutions where 2t = π/6, 5π/6, 13π/6, or 17π/6 (adding 2π for the second period). This gives t = π/12, 5π/12, 13π/12, and 17π/12. Option A only provides two solutions, missing that with the coefficient 2 inside the sine, we get two complete periods in [0,2π]. When the argument has a coefficient greater than 1, expect more solutions in the given interval.

This problem asks us to solve a trigonometric equation in the context of a Ferris wheel's height. The key is to isolate the sine function by setting h(t) = 25, which gives us 15sin(πt/20) + 18 = 25, so sin(πt/20) = 7/15. Using inverse sine, we get πt/20 = arcsin(7/15) ≈ 0.486 radians, giving t ≈ 9.95 seconds. Since sine is positive in both the first and second quadrants, we also have πt/20 = π - 0.486 ≈ 2.656 radians, giving t ≈ 30.05 seconds. Option B incorrectly gives only one solution, missing that sine equations typically have two solutions per period. To solve trigonometric equations in context, always isolate the trig function first, then consider all solutions within the given domain.

This is solving a trigonometric equation for a mass-spring system's position. To find when y(t) = 2, we solve 4cos(3t) = 2, which gives cos(3t) = 0.5. Using inverse cosine, we get 3t = π/3, so t = π/9 seconds for the first occurrence. Option B suggests t = π/6, but substituting gives cos(π/2) = 0, not the required 0.5. Option C gives t = π/3, which yields cos(π) = -1, also incorrect. When solving trigonometric equations with a coefficient inside the argument, always divide by that coefficient after applying the inverse function to find the time value.

This problem involves solving a trigonometric equation for daylight hours in a yearly cycle. Setting D(d) = 13.5, we get 3sin(2π(d-80)/365) + 12 = 13.5, so sin(2π(d-80)/365) = 0.5. The sine equals 0.5 at π/6 and 5π/6 radians. For the first solution: 2π(d-80)/365 = π/6 gives d ≈ 111 days. For the second: 2π(d-80)/365 = 5π/6 gives d ≈ 232 days. Option B's values of 49 and 294 would place the solutions too early and too late in the year respectively. When solving periodic models over a year, ensure your solutions fall within the realistic range of 1 to 365 days.