Question

How do you solve `cos x - 3 cos (x/2) = 0`?

Answer 1

`x=212'37'`

Explanation:

`cosx-3cos(x/2)=0`

`cos(x/2times2)-3cos(x/2)=0`

Double angle formula `cos2theta=cos^2theta-sin^2theta`
`cos^2(x/2)-sin^2(x/2)-3cos(x/2)=0`

Rearrange so that `cos^2theta-sin^2theta=cos^2theta-(1-cos^2theta)=cos^2theta-1+cos^2theta=2cos^2theta-1`
`2cos^2(x/2)-3cos(x/2)-1=0`

Using quadratic formula:

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`cos(x/2)=(3+-sqrt(9-4(2)(-1)))/4`

`cos(x/2)=(3+-sqrt17)/4`

`x/2=cos^(-1)((3+-sqrt17)/4)`

`x/2=cos^(-1)((3-sqrt17)/4)` only

Why? Well, `(3+sqrt17)/4` is greater than 1 and since the domain of `cos^(-1)x` is `-1< x<1`, then there will be no solution

Now, if the domain was between `0 < x<360` then it would become `0< x/2<180`

`x/2=106'18'`

`x=212'37'`

Answer 2

`x = 174^@4 + k720^@`.
`x = 212^@60 + k720^@`

Explanation:

`cos x - 3cos (x/2) = 0`
Note: `cos x = 2co^2 (x/2) - 1` (trig identity)
`2cos^2 (x/2) - 3cos (x/2) - 1 = 0`.
Solve this quadratic equation for `cos (x/2)`.
`D = d^2 = b^2 - 4ac = 9 + 8 = 17` --> `d = +- sqrt17`
There are 2 real roots:
`cos (x/2) = -b/(2a) +- d/(2a) = 3/4 +- sqrt17/4`
`cos (x/2) = (3 + sqrt17)/4` (rejected because > 1)
`cos (x/2) = (3 - sqrt17)/4 = - 0.28`
Calculator and unit circle gives 2 solutions for `x/2`.
`x/2 = +- 106^@30`

a. `x/2 = 106^@30 + k360^@`
`x = 212^@60 + k720^@`
b. `x/2 = - 106^@30`, or its co-terminal
`x/2 = 360 - 106.30 = 253^@70 + k360^@`
`x = 507.4 = 507.4 - 360 = 147^@4 + k720^@`