Question

sin(2x)=rad3/2 steps to solve

Answer 1

cos4x=-\[\sqrt{3}\]/2

Answer 2

-rad3/2=cos4x

Answer 3

tan3x=rad3/3

Answer 4

sin(2x/3)=-1

Answer 5

sin(2x+pi/6)=1/2

Answer 6

and tan(x/2)=rad 3

Answer 7

-rad(3/2)=cos4x=1-2sin^2(2x)
\[\sin2x=2\sin^22x-1\]
\[2\sin^22x-sinx-1=2\sin^22x-2\sin2x+\sin2x-1=(2\sin2x+1)(\sin2x-1)\]
\[\sin2x=1,-1/2\]

Answer 8

sin(2x)=sqrt.3/2

2x = arcsin(sqrt.3/2)

x = arcsin(sqrt.3/2)/2

Answer 9

co4x = \[(\sqrt{3})\div2\]

Answer 10

i can only ue identitie to solve and i dont think it uses arcsin

Answer 11

identity for sin and cos is: sin^2+cos^2=1

Answer 12

ye i know all the differentys including the double anlge ones the half angle ones and the power reducing ones

Answer 13

and the identites of:

sin(a+b)=sina cosb+sinb cosa

cos(a+b) = cosa cosb - sina sinb

this tiny little laptop just aint good on for this site :/

Answer 14

ye i know those as well

Answer 15

u can usew those to derive the double angles cause i did that but how do u solve thee i dont get it cause it syas u can have more 4 anwer for eafch

Answer 16

the trigs are called circle function since they circle back around time and again to the same "spot" but with a new "name"

Answer 17

so if you know an angle; the sin and cos parts will also apply for +- multiples of 360 degrees

Answer 18

drawing rt triangles helps at times too

Answer 19

amis i know trig i just dont get how to solve this specific ones, so if u could solve thee then i would get it or atleast the firsst one.

Answer 20

i cant draw with this thing so ill try the first one at least

Answer 21

so for the first one i got pi/3 and pi/6

Answer 22

sin(2x) = sqrt.3/2

sin = over/hyp ... so 2x = 60 degrees

x = 30 degrees as a start

Answer 23

but the book syas su can have 4 answers excluding revolutions

Answer 24

since sin is positive in quads 1 and 2

we have 2x = 60; or 2x = 120

Answer 25

5pi/6

Answer 26

since the periodicity is doubled; we go from a cycle of 2pi,to a cycle of pi

Answer 27

so we get 4 results; 30 60 210 and 330 i believe

plus rotations

Answer 28

maybe