# Find all solutions of the equation in the interval [0, 2π). sin(x/2)+cosx-1=0

Hi! To solve this equation we need to express `cos(x)` in terms of `sin(x/2).` It is a well-known formula: `cos(x) = cos^2(x/2) - sin^2(x/2) = 1 - 2sin^2(x/2).`

Now denote `sin(x/2)` as `u` and obtain an equation for `u:` `u + 1 - 2u^2 - 1 = 0,` or `u...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Hi! To solve this equation we need to express `cos(x)` in terms of `sin(x/2).` It is a well-known formula: `cos(x) = cos^2(x/2) - sin^2(x/2) = 1 - 2sin^2(x/2).`

Now denote `sin(x/2)` as `u` and obtain an equation for `u:` `u + 1 - 2u^2 - 1 = 0,` or `u - 2u^2 = 0,` or `u(1 - 2u) = 0.`

This gives `u = 0` or `u = 1/2.` Now recall that `u = sin(x/2)` and solve for `x` the equations `sin(x/2) = 0` and `sin(x/2) = 1/2.`

There is the only solution for `sin(x/2) = 0` on `[0, 2 pi)` , `x_1 = 0` (`x/2 = k pi` , `x = 2k pi` ).

There are two solutions for `sin(x/2) = 1/2` on `[0, 2 pi)` , `x_2 = pi/3` and `x_3 = (5 pi)/3.`

(`x/2 = pi/6 + 2k pi` or `x/2 = -pi/6 + 2k pi` ).

So the answers are `x_1 = 0` , `x_2 = pi/3` and `x_3 = (5 pi)/3.`

Approved by eNotes Editorial Team