Asked by muzi
solve for x for angles between o and 360 degrees
(i)cos2x+sinx=0
(i)cos2x+sinx=0
Answers
Answered by
Reiny
cos2x+sinx=0
(1 - 2sin^2 x) + sinx = 0
2sin^2 x - sinx - 1 = 0
(2sinx +1)(sinx - 1) = 0
sinx = -1/2 or sinx = 1
for sinx = -1/2, x is in quads III or IV
x = 210° or 330°
for sinx = 1
x = 90°
x = 90° , 210° or 330°
(1 - 2sin^2 x) + sinx = 0
2sin^2 x - sinx - 1 = 0
(2sinx +1)(sinx - 1) = 0
sinx = -1/2 or sinx = 1
for sinx = -1/2, x is in quads III or IV
x = 210° or 330°
for sinx = 1
x = 90°
x = 90° , 210° or 330°
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.