Solve 3sin2x - 1 = 0

Interval: [0, 2 pi]

Solve 6cos^2x + 5cosx - 6 = 0

1 answer

For the first one, sin2x = 1/3
2x = 0.33984, 2.80176 or 6.62302 radians (and one more less than 4 pi)
x = 0.16992, 1.4088, or 3.3115
(and one more less than 2 pi that I will leave you to figure out)

For the second question, factor into
(2cosx +3)(3cosx -2) = 0
Then set each factor = 0 and solve for cos x.
Similar Questions
    1. answers icon 2 answers
  1. Compute inverse functions to four significant digits.cos^2x=3-5cosx rewrite it as.. cos^2x + 5cosx -3=0 now you have a
    1. answers icon 0 answers
  2. Compute inverse functions to four significant digits.cos^2x=3-5cosx cos^2x + 5cosx -3=0 now you have a quadratic, solve for cos
    1. answers icon 0 answers
  3. i have two questions for you to check please!6cos^2x+5cosx-4=0 where 0degrees < or equal to 0 which is < or equal to 360. I got
    1. answers icon 0 answers
more similar questions