Simple Trig Equations Help (1 Viewer)

[xXmuffin0manXx]

Solve the following trig equations.

i and ii are equal to or greater than
(<_ = equal to or greater than)

i) 2sinx - root3 = 0 for x <_ 0 <_ 360

ii) root3 tanx + 1 for -360<_ x <_ 360

iii) 2cosx = - root3 for -180 x < x < 180

_______________________________

Solve for -180 <_ x <_ 180

i) cosx = 0.187

ii) 3(tan^2)x = 1

iii) 4sinx + 3 = 0

__________________________________

i) Solve sin2x = 3/5 for 0 <_ x <_ 360



thanks and working please!!

 

[Timothy.Siu]

these are really really basic questions, i suggest u read examples and you should be fine.

i'll do the first one

i) 2sinx - root3 = 0 for x <_ 0 <_ 360
sin x=root3/2
x=60 or 120

i) cosx = 0.187
x=79.22 (2dp) or 280.78 (dp)
note that cos (360-x)=cos x

 

[Miss Sunshine]

muffinman is this yr 10 or 11 trig?

i recall some questions from year 10

 

[gurmies]

ii) root3 tanx + 1 for -360<_ x <_ 360

I'm assuming that's = 0.

root3 tanx + 1 = 0

root3tanx = -1

tanx = -1/root3

x = 150, 330, -30, -210

iii) 2cosx = - root3 for -180 x < x < 180

cosx = -root3/2

x = 150, -150

Solve for -180 <_ x <_ 180:

ii) 3(tan^2)x = 1

(tan^2)x = 1/3

tanx = +- 1/root3

x = 30, 150, -30, -150

iii) 4sinx + 3 = 0

sinx = -3/4

x = -48'35" , -131'25"

Solve sin2x = 3/5 for 0 <_ x <_ 360

0 <_2x <_ 720

2x = 36'52", 143"8', 396'52", 503"8'

x = 18' 26", 71' 34", 198' 26", 251' 34"

 

[kurt.physics]

Solve the following trig equations.

i and ii are equal to or greater than
(<_ = equal to or greater than)

i) 2sinx - root3 = 0 for x <_ 0 <_ 360

ii) root3 tanx + 1 for -360<_ x <_ 360

iii) 2cosx = - root3 for -180 x < x < 180

_______________________________

Solve for -180 <_ x <_ 180

i) cosx = 0.187

ii) 3(tan^2)x = 1

iii) 4sinx + 3 = 0

__________________________________

i) Solve sin2x = 3/5 for 0 <_ x <_ 360



thanks and working please!!

i)2sinx - root3 = 0 for x <_ 0 <_ 360

2sin x = root3
sin x = root3 / 2

x = 60, 120

ii)root3 tanx + 1 for -360<_ x <_ 360

what does it equal?

iii)2cosx = - root3 for -180 < x < 180

cos x = -root3/2

cos x = - (root3/2)

the angle whose cosine gives root3/2 is 30 degrees

cosine is negative in second quad and 3rd quad.

Hence x = 180 - 30 and -180 + 30

x = 150, -150

Second group

i)cosx = 0.187

x = 72'20", -72'20"

ii)3(tan^2)x = 1

(tanx)² = 1/3

tan x = +- (1/root3)

x = 30', -30' , 150', -150'

iii)4sinx + 3 = 0

4sinx = -3

sinx = -3/4

the angle whose sine is 3/4 is 48' 35"

so x = -48' 35", -131' 25"


Solve sin2x = 3/5 for 0 <_ x <_ 360

the range is 0 <_ 2x <_ 720

sin^-1(3/5) = 36' 52"

Hence 2x = 36' 52"

sine is positive in the first and second quadrants

2x = 36' 52", 143' 8", 396' 52", 503' 8"

x = 18'26" , 71'34" , 198' 26" , 251' 34"

Any questions

 

[kurt.physics]

ii) root3 tanx + 1 for -360<_ x <_ 360

I'm assuming that's = 0.

root3 tanx + 1 = 0

root3tanx = -1

tanx = -1/root3

x = 150, 330, -30, -210

iii) 2cosx = - root3 for -180 x < x < 180

cosx = -root3/2

x = 150, -150

Solve for -180 <_ x <_ 180:

ii) 3(tan^2)x = 1

(tan^2)x = 1/3

tanx = +- 1/root3

x = 30, 150, -30, -150

iii) 4sinx + 3 = 0

sinx = -3/4

x = -48'35" , -131'25"

Solve sin2x = 3/5 for 0 <_ x <_ 360

0 <_2x <_ 720

2x = 36'52", 143"8', 396'52", 503"8'

x = 18' 26", 71' 34", 198' 26", 251' 34"

Darn! Beaten to it! ;P

 

[gurmies]

haha, sorry man, if I had known you were in the process of writing it, I would have left it to you

 

[kurt.physics]

haha, sorry man, if I had known you were in the process of writing it, I would have left it to you

No, its cool. In fact, two people getting the same results just vindicates the results more

 

[gurmies]

^ Agreed.

 

[xXmuffin0manXx]

i)2sinx - root3 = 0 for x <_ 0 <_ 360

2sin x = root3
sin x = root3 / 2

x = 60, 120

ii)root3 tanx + 1 for -360<_ x <_ 360

what does it equal?

iii)2cosx = - root3 for -180 < x < 180

cos x = -root3/2

cos x = - (root3/2)

the angle whose cosine gives root3/2 is 30 degrees

cosine is negative in second quad and 3rd quad.

Hence x = 180 - 30 and -180 + 30

x = 150, -150

Second group

i)cosx = 0.187

x = 72'20", -72'20"

ii)3(tan^2)x = 1

(tanx)² = 1/3

tan x = +- (1/root3)

x = 30', -30' , 150', -150'

iii)4sinx + 3 = 0

4sinx = -3

sinx = -3/4

the angle whose sine is 3/4 is 48' 35"

so x = -48' 35", -131' 25"


Solve sin2x = 3/5 for 0 <_ x <_ 360

the range is 0 <_ 2x <_ 720

sin^-1(3/5) = 36' 52"

Hence 2x = 36' 52"

sine is positive in the first and second quadrants

2x = 36' 52", 143' 8", 396' 52", 503' 8"

x = 18'26" , 71'34" , 198' 26" , 251' 34"

Any questions

how did you get -210 and -30 for ii) ??

 

[suzlee]

how did you get -210 and -30 for ii) ??

cos the question says -360<_x<_360 you have to include 2 revolutions

0-30=-30 and -180-30=-210

 

[suzlee]

i) cosx = 0.187
x=79.22 (2dp) or 280.78 (dp)
note that cos (360-x)=cos x

but it says -180 <_ x <_ 180 so shouldn't it be 79', -79'? :S

 

[Concerti]

준호야..
어디서 이런 Q's 을난거야..?
matrix..?

haha

 

[kaz1]

준호야..
어디서 이런 Q's 을난거야..?
matrix..?

haha

What's with the Korean? You get these types of questions from most textbooks, so it doesn't necessarily mean he goes to matrix.