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4 Trig Eq. Q's (1 Viewer)

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i) Solve 2cos3x = -1 for -90 <_ x <_ 90

ii) tan x/2 = -2.6 for -360 <_ x <_360

iii) 4cos2x = 3sin2x for x <_ x <_ 180

iv) Solve 4cos^2 x = 3 for 0 <_ x <_ 360


Thanks
Working too please
 

kaz1

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i) 2cos3x = -1
cos3x=-1/2
3x=±120,±240
x=±40,±80
iii)4cos2x = 3sin2x
4=3tan2x
tan2x=4/3
2x=53o8oo,233o8oo
x=26o34oo,116o34oo
 
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lolman12567

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xXmuffin0manXx said:
iv) Solve 4cos^2 x = 3 for 0 <_ x <_ 360
iv) cos2x = 3/4

cosx = + or - root3/2

.'. x = 30, 150, 210, 330

hopefully this is right
 

bored of sc

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xXmuffin0manXx said:
ii) tan x/2 = -2.6 for -360 <_ x <_360
tan(x/2) = -2.6
x/2 = tan-1(-2.6)
= -68.96248897o

-180 < x/2 < 180
x/2 = -68.96248897o, 111.037511o

-360 < x < 360
x = -137.9249779o, 222.0750221o
x = -137.9o, 222.1o (to 1.d.p)
 

Michaelmoo

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xXmuffin0manXx said:
How did you get the mathematical symbols?

And thanks all for the help!
The greater than equal to is just > underlined like this >
 

h3ll h0und

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can someone solve :

(sin^2 (x)- cos^2 (x)).(1-sinxcosx)/cosx(secx-cosecx)(sin^3(x)+cos^3(x))=sinx

i expanded the difference of the two squares and also (sin^3(x)+cos^3(x)) and then crossed out and was left over with sinx-cosx/cosx(secx-cosecx).....leading me to sinx-cosx/1-cotx ... now what?
 
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Timothy.Siu

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h3ll h0und said:
can someone solve :

(sin^2 (x)- cos^2 (x)).(1-sinxcosx)/cosx(secx-cosecx)(sin^3(x)+cos^3(x))=sinx

i expanded the difference of the two squares and also (sin^3(x)+cos^3(x)) and then crossed out and was left over with sinx-cosx/cosx(secx-cosecx).....leading me to sinx-cosx/1-cotx ... now what?
u can change the cot to 1/tan
and then u get (sinxtanx-cosxtanx)/(tanx-1)=sinx(tanx-1)/tanx-1=sin x=RHS

note:cosxtanx=sinx
 

kaz1

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Top bit can be

(-cos22x)(1-(1/2)sin2x)
(sin22x-1)(1-(1/2)sin2x)
 

h3ll h0und

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OOH ....i got it thanx....lol i avoid turning it into a fraction at the bottom...i dunno y.... it makes it look more confusing ...i guess ill get used to it though xD
 

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