Question #265f1

2 Answers
Feb 14, 2018

pi/3 or {5pi}/3

Explanation:

Use the double angle formula
cos (2A) = 1-2sin^2 A
to rewrite the equation in the form
0 = sin(x/2)- cos(x) = sin(x/2)-1+2sin^2(x/2)

Substituting t = sin(x/2) this equation becomes
2t^2+t-1 = 0 implies (2t-1)(t+1) = 0

Since x in [0,2pi), we have x/2 in [0,pi), so that t = sin(x/2) >= 0 Thus
2t-1 = 0 implies t=1/2 implies sin(x/2) = 1/2
Thus, either x/2 = pi/6 or x/2 = {5pi}/6

Feb 14, 2018

x=+-pi or x=+-pi/2

Explanation:

Starting with sin(x/2)-cosx=0, use the half angle formula for sin(x/2):
sin(x/2)=sqrt((1-cosx)/2) and rewrite the original equation:

sqrt((1-cosx)/2)-cosx=0
Add cosx to both sides: #sqrt((1-cosx)/2)-cosx=0 =>

sqrt((1-cosx)/2)=cosx
Square both sides to obtain (1-cosx)/2=cos^2x
Multiply both sides by 2, then put everything on one side:
1-cosx=2cos^2x
2cos^2x +cosx-1=0
You've essentially got a simple quadratic equation to factor now. If n=cosx, you could rewrite it as 2cos^2x -cosx+1=0
=> 2n^2-n-1, which factors as (2n-1)(n+1).

So 2cos^2x +cosx-1=0 => (2cosx -1)(cosx+1)=0

Like solving for the zeros in any quadratic equation, set each factor equal to zero to find the solutions:
2cosx-1=0
2cosx=1

We know from the unit circle that cosx=1/2 at x=+-pi/3, or equivalently, x=+-60^o

The next factor:
cosx+1=0
cosx=-1

Likewise, the unit circle shows us that cosx=-1 where x=+-pi, and that +-pi=+-180^o

So, x=+-pi or x=+-pi/3
Or in degrees: x=+-180^o or x=60^o