Question

Solve the equation on the interval 0≤ θ <2 π. 6 sin 2 θ + 13 sin θ + 7 = 0

Answer 1

To solve the equation 6sin^2(θ) + 13sin(θ) + 7 = 0 on the interval 0≤ θ < 2π, we can use the quadratic formula.

Step-by-step explanation:

To solve the equation 6sin^2(θ) + 13sin(θ) + 7 = 0 on the interval 0≤ θ < 2π, we can use the quadratic formula. Let's treat sin(θ) as a variable and rewrite the equation:

6(sin^2(θ)) + 13sin(θ) + 7 = 0

Now, let's substitute x for sin(θ) and solve the quadratic equation:

6x^2 + 13x + 7 = 0

Using the quadratic formula, we have:

x = (-b ± √(b^2 - 4ac))/(2a)

Substituting the values a=6, b=13, and c=7 into the formula:

x = (-13 ± √(13^2 - 4(6)(7))) / (2(6))

Now, we can solve for x and find the values of sin(θ) that satisfy the equation.