How do you solve sinx = 1 - cosx over the interval (0, 2pi]?

sinx = 1 - cosx solve over the interval [0,2pi]

1 Answer
Feb 15, 2018

The final solutions are x=2pik,pi/2+2pik.

Explanation:

Use the modified Pythagorean identity:

sin^2x+cos^2x=1

=>sin^2x=1-cos^2x

First, square both sides of the equation:

sinx = 1 - cosx

sin^2x=(1-cosx)^2

sin^2x=1-2cosx+cos^2x

0=1-2cosx+cos^2x-sin^2x

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

0=1-2cosx+cos^2x-1+cos^2x

0=1-1-2cosx+2cos^2x

0=2cos^2x-2cosx

0=cos^2x-cosx

0=cosx(cosx-1)

cosx=0, cosx=1

The solutions for cosx=0 are pi/2 and (3pi)/2, and the solution for cosx=1 is 0.

Now, plug all these values back into the original equation to make sure they hold true:

sin(pi/2) stackrel(?)= 1-cos(pi/2)

1stackrel(?)=1-0
1=1

The solution pi/2 works.

sin((3pi)/2)stackrel(?)=1-cos((3pi)/2)

-1stackrel(?)=1-0
-1!=1

The solution (3pi)/2 doesn't work.

sin(0)stackrel(?)=1-cos(0)

0stackrel(?)=1-1

0=0

The solution 0 works.

The final solutions are pi/2 and 0. Now, we have to add 2pik to each of them, to represent that the solution would still be valid after any full rotation (k represents any integer). The full answer is now:

x=0+2pik,pi/2+2pik