A community for students.
Here's the question you clicked on:
← 55 members online
 0 viewing
anonymous
 4 years ago
solve:
sin^2x=sinxcosx
anonymous
 4 years ago
solve: sin^2x=sinxcosx

This Question is Closed

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.4sin^2x =sin x cosx sin^2 xsinx cos x=0 sinx (sin xcos x)= either sin x=0 or sin x = cos x sin x =0 so \[x= n \pi\] sin x = cos x \[x= 2n \pi + \pi/4\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The answer is x=0, pi over 4, pi, 5pi over 4. and how would I get 5pi over 4? Thanks.

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.4yeah sorry,it'll also include \[n \pi+ \pi/4\]

ash2326
 4 years ago
Best ResponseYou've already chosen the best response.4so you'll get, 0 , pi/4 and 5pi/4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think I got it. Thanks.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.