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.
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.
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
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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.
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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!