An equation on the form has the solutions and
An equation on the form has the solutions and
In your case: or
Okay, I always struggle with how to exactly work out the algebra for these trigonometric problems... The question is to identify where and what type of critical points lie on the function:
between 0 and 2pi. Apparently there are 4.
So far this is what I have done:
First of all I don't know how to solve for x in the final line without a calculator. Also, I only see that that would only possibly lead to two critical points between 0 and 2pi....
Thank you anyone who can help me a bit!
P.S. I kind of suck at this math formatting so forgive me...
Is that really what the problem said? That "1" in front of the sin(x) doesn't add anything and is peculiar.
Ifbetween 0 and 2pi. Apparently there are 4.
So far this is what I have done:
As Mondreus told you, either cos(x)= 0, which happens at and at or sin(x)= -1/4 which happens at x= 3.3943 and at x= 6.0305, approximately. There is no "simple" form for that. Why do you want to solve it "without a calculator"?
First of all I don't know how to solve for x in the final line without a calculator. Also, I only see that that would only possibly lead to two critical points between 0 and 2pi....
Thank you anyone who can help me a bit!
P.S. I kind of suck at this math formatting so forgive me...
I got the solutions now.... and yeah, the best I could do for sinx=-1/4 was 2pi+arcsin(-1/4) so I was wondering if I was missing out. In my calculus calculators are not permitted, in any of the quizzes or final exam, so I thought this was a strange solution.
And whoever pointed out the "1" in front of sinx I know that is very peculiar, I don't know why I decided to copy it directly from the question, but that's definitely how it appeared.
I found a video and then skeeter explained that you must factor in order to not lose answers, so that's all good.
Thank you everyone!