> #trigonometry , #equations > Solve the following trigonometric equations for 0°≤∆≤360° , 8sin²∆+2sin∆-1=0?
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more than 2 weeks ago

Solve the following trigonometric equations for 0°≤∆≤360° , 8sin²∆+2sin∆-1=0?

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more than 2 weeks ago
To solve the equation 8sin²∆ + 2sin∆ - 1 = 0, we can use a substitution to simplify the equation.

Let's substitute sin∆ with x. The equation becomes:

8x² + 2x - 1 = 0

Now we can solve this quadratic equation for x.

Using the quadratic formula, x = (-b ± √(b² - 4ac)) / (2a), where a = 8, b = 2, and c = -1.

x = (-2 ± √(2² - 4(8)(-1))) / (2(8))
x = (-2 ± √(4 + 32)) / 16
x = (-2 ± √36) / 16
x = (-2 ± 6) / 16

There are two possible solutions for x:

1) x = (6 - 2) / 16 = 4 / 16 = 1/4
2) x = (-6 - 2) / 16 = -8 / 16 = -1/2

Now we need to find the corresponding values of ∆ for these solutions.

1) If sin∆ = 1/4, then ∆ = sin^(-1)(1/4) ≈ 14.48° or ∆ = 180° - 14.48° ≈ 165.52°

2) If sin∆ = -1/2, then ∆ = sin^(-1)(-1/2) ≈ -30° or ∆ = 180° + 30° ≈ 210°

Therefore, the solutions for 0° ≤ ∆ ≤ 360° are ∆ ≈ 14.48°, ∆ ≈ 165.52°, ∆ ≈ -30°, and ∆ ≈ 210°.
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