The expression lie between

Question

5cosΘ+3cos(Θ+pi/3)+3

 

Answer 1

cos(x+pi/3) = cosxcospi/3 - sinxsinpi/3 = (1/2)cosx - (sqr3/2)sinx

5cosx + (3/2)cosx - (3sqr3/2)sinx = 3

(13/2)cosx - (3/2)sqr3sinx = 3

sinx =sqr(1-cos^2x)

(13/2)cosx - (3/2)sqr3sqr(1-cos^2x) = 3

(13/2)cosx - 3 = (3/2)sqr3sqr(1-cos^2x)

square both sides

(169/4)cos^2x - 39cosx + 9 = (27/4)(1-cos^2x)

(169/4 - 27/4)cos^2x -39cosx + 9-27/4 = 0

(71/2)cos^2x - 39cosx +9/4 = 0

use the quadratic formula to solve for cosx

cosx = 39/71 + or - (1/71)sqr(39^2 - 4(71/2)(9/4))

then use a calculator with an inverse trig function

x = about 86.5 degrees