Question

Solve on interval 0,2pi

Answer 1

\[(cosx+1)(2\cos^2 x-3cosx-2)\]

Answer 2

ok hold on

Answer 3

there's no equation to solve :O
what you've written is an expression,
is the Q by any chance,
\((cosx+1)(2\cos^2 x-3cosx-2)=0\)
?

Answer 4

\[x=2\pi,x=\frac{ \pi }{ 2 },x=\frac{ \pi }{ 3 }\]
\[x=\frac{ \pi }{ 6 },x=\frac{ 7\pi }{ 6 }\]
\[x=\pi,x=\frac{ 2\pi }{ 3 },x=\frac{ 4\pi }{ 3 }\]
\[x=2\pi,x=\frac{ \pi }{ 2 },x=\frac{ 5\pi }{ 4 }\]

Answer 5

no sorry thats all i have

Answer 6

@hartnn help?

Answer 7

its this only : \((cosx+1)(2\cos^2 x-3cosx-2)=0\)
so, we have \((cosx+1)=0 \quad and \quad (2\cos^2 x-3cosx-2)=0\)
the first one is easy, any ideas what to do for \( (2\cos^2 x-3cosx-2)=0\)
?

Answer 8

simplifies to
\[(2cosx+1)(cosx-2)\]
?

Answer 9

graph it lol.

Answer 10

yes, \((2cosx+1)(cosx-2)=0\)
is correct.
now just solve these,
\((2cosx+1)=0 \quad \\ (cosx-2)=0\)

Answer 11

how would i sole them?

Answer 12

2 cos x +1 =0
isolate cos x, can u ?

Answer 13

like for
cos x -2 = 0
cos x = 2

Answer 14

\[cosx=-\frac{ 1 }{ 2 }\]?

Answer 15

thats correct, now for which values of x, is cos x = -1/2 ?

Answer 16

what do you mean?
sorry ive been up all night!!!

Answer 17

i mean there are 2 angles for which cos x equals -1/2
like, 2 values for which cos x =+1/2 will be pi/3 and 5pi/3

Answer 18

also, cos x = 2 is not possible because cos x can take values from +1 to -1 only.

Answer 19

how would i find the radical?

Answer 20

from the unit circle, you can find that , cos x = -1/2 for x =2pi/3 and 4pi/3

Answer 21

okay

Answer 22

and cos x +1 =0 would give you
cos x = -1
you know for which angle is cos x = -1 ??

Answer 23

do you have an example of a unit circle? i found one online but its confusing

Answer 24

http://en.wikipedia.org/wiki/File:Unit_circle_angles_color.svg

Answer 25

you just look where, x-cordinate (cos) equals, -1/2
if you search, you will find it at 2 point, x = 2pi/3 and x= 4pi/3
now search where x-co-ordinate is -1 ?

Answer 26

\[\pi,\frac{ 3\pi }{ 2 }?\]

Answer 27

not at 3pi/2
at 3pi/2, x-cordinate = 0
so, only x= pi

Answer 28

oops sorry

Answer 29

that means cos pi=-1

Answer 30

now we have only 3 roots , x=pi, 2pi/3,4pi/3

Answer 31

so it would be C?

Answer 32

yes :)
any doubts ?

Answer 33

Yay! you're so smart! :)

Answer 34

you too are smart enough :)

Answer 35

lol smart enough! :) gee thanks!

Answer 36

welcome ^_^