Basic Trig Question (1 Viewer)

[GaDaMIt]

Solve for 0 =< x =< 360

root 3 sin x + cos x = 0


show working or explain how you did it or something please

 

[Elvin]

don't use t formula.

 

[webby234]

You don't need t, you've missed the simple solution!

rt3 sinx = -cosx

-1/rt3 = sinx/cosx

tanx = -1/rt3

x = 150, 330

 

[housemouse]

T results are for 3 unit

 

[Elvin]

hence the don't use t formula.

 

[GaDaMIt]

ok thanks for that.. just one more..

Solve each of these homogeneous equations for 0 =< x =< 360 by dividing both sides by a suitable power of cos x. Give solutions to the nearest minute where necessary.

sin^3x + 2 sin^2x cosx + sinx cos^2x = 0

answers say 0, 180, 360, 135, 315.. im only getting those last two.. 135, 315?

i get up to (tanx + 1)^2 = 0..

 

[Trev]

Taking out a factor of sinx and dividing by cos²x you should get:
sinx(tan²x+2tanx+1)=0
sinx(tan²x+1)²=0
The sinx will give you the other required solutions.

 

[webby234]

sin^3x + 2 sin^2x cosx + sinx cos^2x = 0

tan³x + 2tan²x + tanx = 0
tanx(tan²x + 2 tanx + 1) = 0
tanx(tanx + 1)² = 0

I think your problem comes here - you are neglecting the possibility that tan x = 0. The solutions are

tanx = -1 OR tanx = 0
So you get 135, 315 for the first one and 0, 180, 360 for the second one.

 

[webby234]

Haha three posts in a minute all giving slightly differnet solutions. airie, that is not the problem, as cos x is only zero at 90, 270 which are not solutions in this case. If cosx is zero, then sinx will still not be zero, so the equation is not satisfied.

 

[Riviet]

Webby's got it right.

 

[GaDaMIt]

Taking out a factor of sinx and dividing by cos²x you should get:
sinx(tan²x+2tanx+1)=0
sinx(tan²x+1)²=0
The sinx will give you the other required solutions.

sin^3x + 2 sin^2x cosx + sinx cos^2x = 0

tan³x + 2tan²x + tanx = 0
tanx(tan²x + 2 tanx + 1) = 0
tanx(tanx + 1)² = 0

I think your problem comes here - you are neglecting the possibility that tan x = 0. The solutions are

tanx = -1 OR tanx = 0
So you get 135, 315 for the first one and 0, 180, 360 for the second one.

Ahh yeah.. i see LOL stupid mistake.. my bad.. thanks HEAPS GUYS !!

 

[airie]

Haha three posts in a minute all giving slightly differnet solutions. airie, that is not the problem, as cos x is only zero at 90, 270 which are not solutions in this case. If cosx is zero, then sinx will still not be zero, so the equation is not satisfied.

Oh yes, that's right Post deleted

 

[Elvin]

Taking out a factor of sinx and dividing by cos²x you should get:
sinx(tan²x+2tanx+1)=0
sinx(tan²x+1)²=0
The sinx will give you the other required solutions.

touche by taking the sinx and not the tanx out.
touche salesman.