Question

Solve the following for θ, in radians, where 0≤θ<2π.
−6sin2(θ)−5sin(θ)+3=0

Answer 1

Answer:correct answers 0.42

2.73

Step-by-step explanation:We can solve this quadratic equation in sin(θ) by using the substitution u = sin(θ):

-6u^2 - 5u + 3 = 0

Now we can use the quadratic formula to solve for u:

u = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = -6, b = -5, and c = 3. Substituting these values, we get:

u = (5 ± sqrt(5^2 - 4(-6)(3))) / 2(-6)

u = (5 ± sqrt(109)) / (-12)

Therefore, either:

sin(θ) = (5 + sqrt(109)) / (-12)

or:

sin(θ) = (5 - sqrt(109)) / (-12)