Question

Question #fcb2e

Answer 1

`xapprox0.251`

Explanation:

`cosxcos3x=0.707`

`f(x)=cosxcos3x-0.707=0`

Use Newton method to find a zero of `f`

`x_(n+1)=x_n-(f(x_n))/(f'(x_n))`

`f'(x)=-2sin2x-2sin4x`

`0.707=cos(1/4pi)`, so a good `x_0` would be `(1/4pi)/3=0.262`

`x_1=0.262-(cos(0.262)cos(3(0.262))-0.707)/(2sin(2(0.262)-2sin(4(0.262))=0.253`

`x_2=0.251`

`x_3=0.251`

`x_4=0.251`

The iterations approach `0.251` to `0.251` is a solution to `cosxcos3x=0.707`

Answer 2

`x = +- 14^@48`

Explanation:

cos x.cos 3x = 0.707
Use trig identity:
`cos a.cos b = (1/2)(cos (a - b) + cos (a + b))`
In this case:
`cos x.cos 3x = (1/2)(cos 2 x + cos 4x)`-->
cos 2x + cos 4x = 2(0.707) = 1.414
Call 2x = X,
cos X + cos 2X - 1.414 = 0.
Replace cos 2X by (2cos^2 X - 1):
`cos X + (2cos^2 X - 1) - 1.414 = 0`
2cos^2 X + cos X - 2.414 = 0.
Solve this quadratic equation for cos X.
`D = d^2 = b^2 - 4ac = 1 + 19.28 = 20.28` --> `d = +- 4.50`
There are 2 real roots:
`cos X = -b/(2a) +- d/(2a) = - 1/4 +- (4.50)/4 = - 0.25 +- 1.125`
a. cos X = - 1.375 (rejected as < - 1)
b. cos 2x = cos X = 0.875
Calculator and unit circle give:
`2x = +- 28^@96` --> `x = +- 14^@48`
Check by calculator:
`x = 14^@48` --> `cos x = 0.97` --> `3x = 43^@44` --> `cos 3x = 0.73`.
cos x.cos 3x = 0.97(0.73) = 0.71. Proved.