Question

6sin2x – 19 sin x + 8 = 0

Answer 1

$y = 6 \sin(2x) - 19 \sin(x) + 8 \\ y = 0$

Answer 2

To solve the equation 6sin^2(x) - 19sin(x) + 8 = 0, we can use some algebraic techniques. Let's solve it step by step:

1. Let's make a substitution to simplify the equation. Replace sin(x) with y, so the equation becomes: 6y^2 - 19y + 8 = 0

2. Now, we need to factor the quadratic equation. It's easier to factor it like this: (2y - 1)(3y - 8) = 0

3. Set each factor to zero and solve for y: 2y - 1 = 0 --> y = 1/2 3y - 8 = 0 --> y = 8/3

4. Since y = sin(x), we have two possibilities: a) sin(x) = 1/2 b) sin(x) = 8/3

However, sin(x) cannot be equal to 8/3 since the sine function's range is -1 to 1.

5. Solve for x when sin(x) = 1/2: x = arcsin(1/2)

Using the unit circle or trigonometric values, we know that arcsin(1/2) = π/6 or 30 degrees.

So the solutions to the equation are: x = π/6 + 2πn (where n is an integer) - in radians x = 30° + 360°n (where n is an integer) - in degrees

These are the values of x that satisfy the original equation.