Asked by sami
Solve 6 sin^(2)x - 5 cos x - 2 = 0 in the interval 0 ≤ x ≤ 2pi
Answers
Answered by
oobleck
since sin^2x = 1-cos^2x, that means
6(1-cos^2x) - 5cosx - 2 = 0
6 - 6cos^2x - 5cosx - 2 = 0
6cos^2x + 5cosx - 4 = 0
(3cosx+4)(2cosx-1) = 0
so, where do you have
cosx = -4/3 ? nowhere
where is cosx = 1/2 ?
6(1-cos^2x) - 5cosx - 2 = 0
6 - 6cos^2x - 5cosx - 2 = 0
6cos^2x + 5cosx - 4 = 0
(3cosx+4)(2cosx-1) = 0
so, where do you have
cosx = -4/3 ? nowhere
where is cosx = 1/2 ?
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.