Asked by Miu Lily
Solve each equation for 0<_ 0<_ 2pi(3.14)
E) 18sin^2(0) + 3cos0 -17 =0
E) 18sin^2(0) + 3cos0 -17 =0
Answers
Answered by
Steve
18sin^2x + 3cosx - 17 0
18 - 18cos^2x + 3cosx - 17 = 0
18cos^2x - 3cosx - 1 = 0
(6cosx-1)(3cosx+1) = 0
cosx = 1/6 or cosx = -1/3
x = 1.40 or 2π-1.40
x = π-1.23 or π+1.23
18 - 18cos^2x + 3cosx - 17 = 0
18cos^2x - 3cosx - 1 = 0
(6cosx-1)(3cosx+1) = 0
cosx = 1/6 or cosx = -1/3
x = 1.40 or 2π-1.40
x = π-1.23 or π+1.23
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.