# core 3 trig (URGENT)

a) Express 3cos x + sin x in the form Rcos(x-W) where R>0 and 0<W<90

6) Using your answers to part (a) or otherwise solve the equation 6cos²x+ sin 2x = 0
for x in the interval o<= x <= 360 giving your answers to 1dp
microwave69
a) Express 3cos x + sin x in the form Rcos(x-W) where R>0 and 0<W<90

Expand Rcos(x-W) and equate coefficients of cosx and sinx with 3cosx+sinx - solve to find R, W.

6) Using your answers to part (a) or otherwise solve the equation 6cos²x+ sin 2x = 0
for x in the interval o<= x <= 360 giving your answers to 1dp

Expand sin2x, factorise out common factors and use (a) when it seems appropriate.
microwave69
a) Express 3cos x + sin x in the form Rcos(x-W) where R>0 and 0<W<90

6) Using your answers to part (a) or otherwise solve the equation 6cos²x+ sin 2x = 0
for x in the interval o<= x <= 360 giving your answers to 1dp

Here is part A. and part B

I don't understand how one could use the solution to part (a). I have the answer.... could someone please explain how I could have used part (a)
microwave69
a) Express 3cos x + sin x in the form Rcos(x-W) where R>0 and 0<W<90

6) Using your answers to part (a) or otherwise solve the equation 6cos²x+ sin 2x = 0
for x in the interval o<= x <= 360 giving your answers to 1dp

In response to Steve's request above:

a.) Let 3cosx + sinx = Rcos(x - W) = R[cosxcosW + sinxsinW] = (RcosW)cosx + (RsinW)sinx

:. 3 = RcosW and 1 = RsinW
-> R^2 = 10 -> R = Sqrt(10)
-> W = sin^-1[1/(Sqrt10)] = 18.4 Deg (1 D.P)

:. 3cosx + sinx = Sqrt10.cos(x - 18.4)

6.) 6cos^2x + sin2x = 0
-> 6cos^2x + 2sinxcosx = 0
-> 2cosx(3cosx + sinx) = 0
-> 2cosx = 0, 3cosx + sinx = 0
-> cosx = 0, cos(x - 18.4) = 0

cosx = 0 -> x = 90, 270 Deg
cos(x - 18.4) = 0 -> x - 18.4 = 90, 270 -> x = 108.4, 288.4 Deg (1.D.P)

:. Solutions for 0 < x < 360 Deg:
x = 90, 108.4, 270, 288.4 Deg (1.D.P)
Nima
In response to Steve's request above:

a.) Let 3cosx + sinx = Rcos(x - W) = R[cosxcosW + sinxsinW] = (RcosW)cosx + (RsinW)sinx

:. 3 = RcosW and 1 = RsinW
-> R^2 = 10 -> R = Sqrt(10)
-> W = sin^-1[1/(Sqrt10)] = 18.4 Deg (1 D.P)

:. 3cosx + sinx = Sqrt10.cos(x - 18.4)

6.) 6cos^2x + sin2x = 0
-> 6cos^2x + 2sinxcosx = 0
-> 2cosx(3cosx + sinx) = 0
-> 2cosx = 0, 3cosx + sinx = 0
-> cosx = 0, cos(x - 18.4) = 0

cosx = 0 -> x = 90, 270 Deg
cos(x - 18.4) = 0 -> x - 18.4 = 90, 270 -> x = 108.4, 288.4 Deg (1.D.P)

:. Solutions for 0 < x < 360 Deg:
x = 90, 108.4, 270, 288.4 Deg (1.D.P)

Thanks Nima, I get it now.
steve2005