Question

Math Analysis Ch. 10.4
Solve the equation for 0 14 years ago

Answer 1

Use trig identity

Answer 2

can you list the steps, i really don't get it :\

Answer 3

would i - the 3sin x over to the other side?

Answer 4

cos(2x) = cos^2(x) - sin^2(x)
= 2 cos^2(x) - 1
= 1 - 2 sin^2(x)

Answer 5

3sin x = 1+ cos 2x
3sin x = 1+ ( 1 - 2 sin^2(x))
2 sin^2(x)+3sin x-2=0
(sinx+2)(2sinx-1)=0
sinx=-2 ,sinx=1/2

Answer 6

okay, ty :) How about number 2?

Answer 7

Change the secx to 1/(cosx)
then change the cos2x to 2 cos^2(x) - 1

Answer 8

then i'd minus 2 cos^2(x) -1 to the other side?

Answer 9

make it like a quadratic and solve by factoring

Answer 10

so i would make 1 have a denominator of cosx so i can be able to add them together?

Answer 11

like this...?
0=1/cosx - 2cos^2x + 1
0=1/cosx - 2cos^2x +cos/cosx
0=1+cosx/cosx - 2cos^2x

Answer 12

cos2x=secx
2cos^2x-1=1/cosx
2cos^3x-cosx=1
cosx-1=0
cosx=1

Answer 13

how did you get 2cos^3x?

Answer 14

wait nvm

Answer 15

so how did you get..
cosx-1=0
cosx=1

Answer 16

factor the cubic equation

Answer 17

so when i get ..
2cos^3-cosx=1
i would minus 1 to the other side again?

Answer 18

yes

Answer 19

why..

Answer 20

2cos^3-cosx=1
2cos^3-cosx-1=0
factor

Answer 21

ok so i got..
cosx(2cos^2x-1)=1

Answer 22

cosx=1
2cos^2x-1=1
+1+1
2cos^2x/2=2/2
square root of(cos^2x)=1
cosx=1..?

Answer 23

so both answers would be cosx=1 o:
RIGHT? :p

Answer 24

dont do 2cos^2x-1=1

Answer 25

use cosx=1 only

Answer 26

why not, that's what i got when i factored...

Answer 27

you cant factor when there's 1 at the right side,
the only thing to factor is right side is zero
2cos^3-cosx-1=0
so you'll get
cosx-1=0
cosx=1

Answer 28

yah, and so i factored
2cos^3x-cosx-1=0
and i got..
cosx(2cos^2x-1)=1

Answer 29

i mean
2cos^3x-cosx=1

Answer 30

like you said an hour ago.

Answer 31

2cos^(3)x-cosx-1=0
Factoring you get
(cosx-1)(2cos^(2)x+2cosx+1)=0

(cosx-1)=0, (2cos^(2)x+2cosx+1)=0
From
cosx-1=0

cosx=1

\[x=\cos^{-1}(1)\]

x=0
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For the quadratic, use quadratic formula
2cos^(2)x+2cosx+1=0

cosx=(-b+-sqrt(b^(2)-4ac))/(2a)

cosx=(-2+-sqrt((2)^(2)-4(2)(1)))/(2(2))

cosx=(-2+-2i)/2(2)

cosx=(-2+-2i)/4

cosx=(-1+i)/2

cosx=(-1-i)/2

cosx=(-1+i)/(2),(-1-i)/2
------------------------------------
cosx=(-1+i)/(2)_cosx=(-1-i)/(2)

cosx=(-1+i)/2 cosx=(-1-i)/2

You don't get any real solution for the quadratic, the only real solution is x=0,180,360

Answer 32

why would x=0,180,and360?

Answer 33

cosx=1 is only on 0 lolz.

Answer 34

0 and 360