anonymous
  • anonymous
Find all solutions on the interval [0,2pi] 2 cosx sinx - cos x =0 Please show+explain work?
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
did you get this yet? if not i will solve.
watchmath
  • watchmath
go ahead :)
anonymous
  • anonymous
first i would write \[2cos(x)sin(x)=cos(x)\] then square to get \[4cos^2(x)sin^2(x)=cos^2(x)\] \[4cos^2(x)[1-cos^2(x)]=cos^2(x)\] \[4cos^2(x)-4cos^4(x)-cos^2(x)=3cos^2(x)-4cos^4(x)=0\]

Looking for something else?

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

More answers

anonymous
  • anonymous
how am i doing so far?
watchmath
  • watchmath
If i do this I will move the cos x to the left and factor out the cos x
anonymous
  • anonymous
\[4cos^4(x) - 3cos^2(x)=0\] \[cos^2(x)(4cos^2(x)-3)=0\] \[cos^2(x)=0\] or \[cos^2(x)=\frac{3}{4}\]
anonymous
  • anonymous
\[cos(x) = 0\] \[x=\frac{\pi}{2}\]or \[x=-\frac{\pi}{2}\]
anonymous
  • anonymous
proof 2 now that i am on a roll. \[f'(x)=3ax^2-4ax-8a\] the zeros are \[\frac{4a\pm\sqrt{16a^2+96a}}{6a}=\frac{4a\pm4\sqrt{a^2+6a}}{6a}=\] \[\frac{2\pm 2\sqrt{a(a+6)}}{3a}\] and in order for this to be -1 \[a(a+6)\] must be a perfect square, a miracle if ever there
anonymous
  • anonymous
ignore that last gibberish it is wrong and not from here anyway.
watchmath
  • watchmath
Alternative solution \(\cos x(2\sin x -1)=0\) \(\cos x=0\Rightarrow x=\pi/2,3\pi/2\) \(\sin x=1/2\Rightarrow x=\pi/6,5\pi/6\).
anonymous
  • anonymous
btw we knew \[\pm\frac{\pi}{2}\]worked to begin with by inspection.
anonymous
  • anonymous
right, much swifter and i retire now. yikes.
anonymous
  • anonymous
Thanks!

Looking for something else?

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