Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
jpjones
Group Title
solve the following equation for x where
0<or = to x<2pi, sin2x=cosx
 one year ago
 one year ago
jpjones Group Title
solve the following equation for x where 0<or = to x<2pi, sin2x=cosx
 one year ago
 one year ago

This Question is Closed

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.3
hint: \(sin2x = 2sinxcosx\)
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.3
well where are you stuck or show me what you did
 one year ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
the equation is not solved
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.3
\[ 2sinxcosx = cosx => 2sinxcosx  cosx = 0 \] why dont you try and solve it
 one year ago

sami21 Group TitleBest ResponseYou've already chosen the best response.1
take cos(x) you will have \[\Large \cos(x)[2\sin(x)1]=0\] now you have two equations \[\Large \implies \cos(x)=0\] \[\Large \implies 2\sin(x)1=0\] solve these two equations.
 one year ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
if i knew how to solve these equations i wouldn;t be asking for help
 one year ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
to this stage i know, but how to solve from here?
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.3
\[\cos(x) = 0 => cosx = \cos\left(\frac{\pi}{2}\right) => x = \frac{\pi}{2} +n*2\pi , n \epsilon\mathbb{Z} \] \[sinx = \sin\left(\frac{\pi}{6}\right) => x = \left(1\right)^{n}*\frac{\pi}{6} +n*\pi , n \epsilon\mathbb{Z} \]
 one year ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
i've been to wolfram alpha to mimi
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.3
lol i didnt use wolfram..
 one year ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
the answer is correct though but i have to show it using the unit circle, or the graph
 one year ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
i think using the sin wave
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.3
dw:1345619287591:dw
 one year ago

sami21 Group TitleBest ResponseYou've already chosen the best response.1
ok let me give you an example then if you have 2cos(x)+1=0 then \[\Large \implies \cos(x)=\frac{1}{2}\] \[\Large x=\cos^{1}(\frac{1}{2})\] there are only two points in 9o.2pi) for which cos(x) is 1/2 they are \[\Large x=\frac{2 \pi}{3}\] \[\Large x=\frac{4 \pi}{3}\] now solve the 2sin(x)1=0 its quite similar to this.
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.3
for sinx dw:1345619404355:dw
 one year ago

sami21 Group TitleBest ResponseYou've already chosen the best response.1
you use wolfram,thats why you are unable to solve these equations by yourself. this wolfram thingy has ruined the students :(. i think at this level students should avoid using that.
 one year ago

Mimi_x3 Group TitleBest ResponseYou've already chosen the best response.3
yeah i agree. you should only use it to check your answers
 one year ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
no i dont use it but i am aware of it
 one year ago

jpjones Group TitleBest ResponseYou've already chosen the best response.0
mimi you've been great i think i should become a fan
 one year ago
See more questions >>>
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.