anonymous
  • anonymous
Solve cos^(2)(x) - 3sin(x) = 3 for x on [0, 2pi] So I have simplified the function down to (-sin(x) - 2)(sin(x) + 1) = 0 or (-sin(x) -1)(sin(x) + 2) = 0 I have determined x = 3pi/2 as sin(3pi/2) = 1/2 Is that the only value possible As I don't think there is a value in the domain of this function that I could use to solve (sin(x) + 2) or (-sin(x) - 2) I could imput arcsin(2) but would that be wrong and if so why?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
watchmath
  • watchmath
because the value of sin is at most 1. SO arcsin(2) is undefined. So the solutions only comes from -sin x -1 =0
anonymous
  • anonymous
sin(3pi/2) = -1
anonymous
  • anonymous
made a typo. so I'm correct.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

watchmath
  • watchmath
yes \(e\pi/2\) is the only answer
anonymous
  • anonymous
so I should always adhere to the range of the unit circle when dealing with trig functions?
watchmath
  • watchmath
I mean \(3\pi/2\)
anonymous
  • anonymous
yeah I was wondering lol

Looking for something else?

Not the answer you are looking for? Search for more explanations.