Curry
  • Curry
solve for x in the interval [o,2pi) cos^2+sin +1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Curry
  • Curry
thiss one is harder and cos4x-cos2x=0
bahrom7893
  • bahrom7893
When in doubt, turn to wolf lol: http://www.wolframalpha.com/input/?i=solve%28cos4x-cos2x%3D0%29
Curry
  • Curry
im not allowed to use a calculator

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More answers

bahrom7893
  • bahrom7893
i honestly don't know how to solve this without one, but likeL Cos4x-Cos2x=0 is easy to see: Cos4x=Cos2x so x=0, cuz Cos0 is 1
Mertsj
  • Mertsj
Are we talking about cos^2xsinx+1?
Curry
  • Curry
well that one but the co2x-cos2x i much harder so that too
Mertsj
  • Mertsj
cos(2x)-cos(2x)=0
Curry
  • Curry
wait not 2x it cos(4x)-cos(2x)=0
Mertsj
  • Mertsj
Let me work on it for a minute.
Mertsj
  • Mertsj
Let 2x=y \[\cos 2y-\cos y=0\]
Mertsj
  • Mertsj
\[\cos 2y=2\cos ^2y-1\]
Curry
  • Curry
that what u poted is another problem
Mertsj
  • Mertsj
\[2\cos ^2y-1-\cos y=0\]
Mertsj
  • Mertsj
\[2\cos ^2y-\cos y-1=0\]
Mertsj
  • Mertsj
\[(2\cos y+1)(\cos y-1)=0\]
Mertsj
  • Mertsj
\[\cos y=-\frac{1}{2}, \cos y=1\]
Mertsj
  • Mertsj
\[y=\frac{2\pi}{3}, \frac{4\pi}{3},0\]
Mertsj
  • Mertsj
But 2x=y so \[2x=\frac{2\pi}{3}, x=\frac{\pi}{3}\]
Curry
  • Curry
wait i dont get hgow u went from cos2y-cosy t cos2y=2cos^2y-1 i mean u get that it is one of the indetities but sshould-1 becausse 2x= y
Mertsj
  • Mertsj
\[2x=\frac{4\pi}{3}, x=\frac{2\pi}{3}\]
Curry
  • Curry
but houldn't it be cos2y=2coss^2x-1
Mertsj
  • Mertsj
|dw:1332296856974:dw|
Curry
  • Curry
ookk kk i get it now
Mertsj
  • Mertsj
Good for you
Curry
  • Curry
so the final answers are 0,pi,pi/3,2pi/3,/34pi/3,5pik
Curry
  • Curry
5pi/3
Curry
  • Curry
3/4pi
Curry
  • Curry
3/4pi
Mertsj
  • Mertsj
|dw:1332297808892:dw|
Curry
  • Curry
shouldn't it be pi also
Mertsj
  • Mertsj
yes
Curry
  • Curry
wait nvm i need help on the first probnlem i poted the cos^2=in=1
Mertsj
  • Mertsj
For the first one, replace cos^2 with 1-sin^2
Curry
  • Curry
kk i got it but what i dont get it back to the previous(1st problem that wqwhat i dont get is y u did 8pi/3 and 1`0 pi/3 likew y did u do a full revolution so like if i wa given another problem hould i do that ame thing to that problem too or is pecial to this problem
Mertsj
  • Mertsj
If you want all the solutions between 0 and 2 pi and you know you're going to be dividing by 2, then you need to find all the solutions in 2 complete revolutions.
Curry
  • Curry
oh i get it now. i substituted like u said i got down to (sin+1)(-ssin+2) but how do u find the angle measures of sin = 2
Mertsj
  • Mertsj
Do you know that the maximum value of the sin function is 1?
Curry
  • Curry
ye
Curry
  • Curry
3pw
Mertsj
  • Mertsj
So sin(x)=2 is impossible and you discard that.
Mertsj
  • Mertsj
What does 3pw mean?
Curry
  • Curry
i mesaying
Curry
  • Curry
and 3pi/2 i the only anwer?
Mertsj
  • Mertsj
yep
Curry
  • Curry
oo kk i get it ty mert i owe u one
Mertsj
  • Mertsj
yw

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