hi. ive been given this equation

12cosx -4 sinx=7 solve between 0<x<360 degrees.

ive already done the harmonic form for f(x)= 12cosx -4 sinx in the form

= Rcos(x + a) and got square root 160cos(x + 18.43)

im just not sure how to solve it

thanks.

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- December 15th 2008, 03:30 AMbrooksySolving trig
hi. ive been given this equation

12cosx -4 sinx=7 solve between 0<x<360 degrees.

ive already done the harmonic form for f(x)= 12cosx -4 sinx in the form

= Rcos(x + a) and got square root 160cos(x + 18.43)

im just not sure how to solve it

thanks. - December 15th 2008, 04:34 AMdifferentiate
Try t method substitution

sin@ = 2t/1 + t^2

cos@ = 1-t^2/1+t^2

12cosx -4 sinx=7 solve between 0<x<360 degrees.

12(1-t^2/1+t^2) - 4(2t/1+t^2) = 7

(12 - 12t^2/1+t^2)- (8t/1 + t^2) = 7

12- 12t^2 - 8t = 7 + 7t^2

12 - 19t^2 - 8t - 7 = 0

5 - 19t^2 - 8t =0

-5 + 19t^2 + 8t = 0

-8 +- sqrt 64 - 4x19 x -5

-------------------------

38

t = 0.344 or -0.765

tan@/2 = 0.344

@/2 = 18 degrees, 59 minutes

@ = 37 degrees, 58 minutes

or 180 + 37.97 = 217degrees, 58 minutes

tan@/2 = 0.765

@/2 37.42

but thingo is negative - second and 4th quadrant

therefore, @/2 = 142.58 or 322.58

@ = 285.16 or 645.16

@ = 285.16 (within the domain)

Rechecking this... It's not that accurate. Sorry, I'm new to trigonometry.

----------------------------------------------

Using the auxiliary angle method:

12cosx -4 sinx=7 solve between 0<x<360 degrees.

acos@ - bsin@ = Acos(@+a)

tana = b/a

tana = 4/12

a = 18 degress, 26 minutes

A = sqrt a^2 + b^2

= sqrt 144 + 16

= sqrt 160

sqrt 160 cos (@ + 18degrees,26 minutes) = 7

cos (@ + 18 degrees, 26 minutes) = 7/sqrt 160

cos 56 degrees, 24 minutes OR 303 degrees, 36 minutes

@ + 18 degrees, 26 mins = 56 deg, 24 mins

@ = 37 degrees, 58 mins

@ + 18 degrees, 26 mins = 303 degrees, 36 mins

@ = 285 degrees, 10 mins

Don't know if its right, but please don't assume it is! - December 15th 2008, 04:37 AMbrooksy
cheers mate , thats brilliant