Question

Question #57960

Answer 1

`cosx+sin(x/2)=0`

`cos(x/2+x/2)+sin(x/2)=0`

`1-2sin^2(x/2)+sin(x/2)=0`

`2sin^2(x/2)-sin(x/2)-1=0`

Let `y=sin^2(x/2)=0`

`2y^2-2y+x-1=0`

`2y(y-1)+1(y-1)=0`

`2y+1=0`
`y-1=0`

`2sin(x/2)+1=0`

`sin(x/2)=-1/2`

`x=360n-60, nin ZZ`

`sin(x/2)=1`

`x=720n+180,ninZZ`

Answer 2

`pi + 4kpi`
`pi/3 + 4kpi`
`(5pi)/3 + 4kpi`

Explanation:

Use trig identity: `cos 2a = 1 -2sin^2 a`
In this case:
`1 - 2sin^2 (x/2) + sin (x/2) = 0`
Bring it to standard form:
`2sin^2 (x/2) - sin x - 1 = 0`
Solve this quadratic equation for `sin (x/2)`:
Since a + b + c = o, use shortcut.
The 2 real roots are: `sin (x/2) = 1` and `sin (x/2) = c/a = - 1/2`
Use trig table and unit circle -->
A. `sin (x/2) = 1` --> `x/2 = pi/2 + 2kpi` --> `x = pi + 4kpi`

B. `sin (x/2) = - 1/2` --> unit circle -->
a. `x/2 = (7pi)/6 + 2kpi` --> `x = (7pi)/3 + 4kpi` or `x = pi/3 + 4kpi`
b. `x/2 = (11pi)/6 + 2kpi` --> `x = (11pi)/3 + 4kpi`, or
`x = (5pi)/3 + 4kpi`