Trigonometry

(7sin(0.5t-π/2)=)/7 5/7 Find the answer to this by solving 7 sin (0.5𝑡 − ) = 5 in the domain 0 < 𝑡 < 3𝜋. 2 You can write your final answer in terms of 𝜋 or round it to 3 decimal places.

Step-by-step solution with explanation

Final Answer

Two solutions: and (rounded to 3 decimal places)

Step-by-step solution

Isolate the sine function

Divide both sides of the equation by 7. This isolates the sine expression so we can work with it directly.

Take inverse sine of both sides

Apply arcsin to both sides. The reference angle is arcsin(5/7) ≈ 0.7754 radians. Sine is positive in Quadrants I and II, so we need two general solutions.

Write both general solutions for sine

When sin(θ) = positive value, θ can be in Quadrant I or Quadrant II. The Quadrant II angle is π minus the reference angle. Here k is any integer.

Solve for t in Quadrant I case

Add π/2 to both sides, then divide by 0.5. This gives the first solution. Check: 0 < 4.692 < 3π ≈ 9.425 ✓

Solve for t in Quadrant II case

Use π − 0.7754 ≈ 2.3662 as the angle, add π/2, then divide by 0.5. Check: 0 < 7.874 < 9.425 ✓

Check next cycle (k=1) for more solutions

Adding one full period (2π/0.5 = 4π ≈ 12.566) to either solution pushes t far beyond 3π ≈ 9.425. So there are exactly two solutions in the given domain.

Understanding this problem

Learning Insight

When you solve sin(θ) = c, the sine function equals the same positive value at TWO angles in every 2π cycle — one in Quadrant I and one in Quadrant II (symmetric about π/2). That's why we always check both cases. The period of sin(0.5t) is 2π/0.5 = 4π, which is why only one cycle fits inside 0 to 3π.

Quick Tip

The two sine solutions in a cycle are always mirror images: θ = arcsin(c) and θ = π − arcsin(c). Write these down first, THEN solve for t. It keeps the algebra clean and you won't miss a solution.

Common Mistake

Students often find only ONE solution (the Quadrant I angle) and forget the Quadrant II solution π − arcsin(c). Always ask: 'Where else is sine positive (or negative)?' before finishing.