Question

solve for x in the interval [o,2pi)
cos^2+sin +1

Answer 1

thiss one is harder
and cos4x-cos2x=0

Answer 2

When in doubt, turn to wolf lol:
http://www.wolframalpha.com/input/?i=solve%28cos4x-cos2x%3D0%29

Answer 3

im not allowed to use a calculator

Answer 4

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

Answer 5

Are we talking about cos^2xsinx+1?

Answer 6

well that one but the co2x-cos2x i much harder so that too

Answer 7

cos(2x)-cos(2x)=0

Answer 8

wait not 2x it cos(4x)-cos(2x)=0

Answer 9

Let me work on it for a minute.

Answer 10

Let 2x=y
\[\cos 2y-\cos y=0\]

Answer 11

\[\cos 2y=2\cos ^2y-1\]

Answer 12

that what u poted is another problem

Answer 13

\[2\cos ^2y-1-\cos y=0\]

Answer 14

\[2\cos ^2y-\cos y-1=0\]

Answer 15

\[(2\cos y+1)(\cos y-1)=0\]

Answer 16

\[\cos y=-\frac{1}{2}, \cos y=1\]

Answer 17

\[y=\frac{2\pi}{3}, \frac{4\pi}{3},0\]

Answer 18

But 2x=y so
\[2x=\frac{2\pi}{3}, x=\frac{\pi}{3}\]

Answer 19

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

Answer 20

\[2x=\frac{4\pi}{3}, x=\frac{2\pi}{3}\]

Answer 21

but houldn't it be cos2y=2coss^2x-1