SOLUTION: solve sin(theta)cos(theta)−sin(theta)−cos(theta)+1=0 also this one sin(theta)+cos(theta)=1 for both theta is greater than or equal to 0 and less than 2 pi

Algebra ->  Trigonometry-basics -> SOLUTION: solve sin(theta)cos(theta)−sin(theta)−cos(theta)+1=0 also this one sin(theta)+cos(theta)=1 for both theta is greater than or equal to 0 and less than 2 pi      Log On


   

Question 927779: solve
sin(theta)cos(theta)−sin(theta)−cos(theta)+1=0
also
this one
sin(theta)+cos(theta)=1
for both
theta is greater than or equal to 0 and less than 2 pi






Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
sin(theta)cos(theta)−sin(theta)−cos(theta)+1=0
sinxcosx-sinx-cosx+1=0
sinx(cosx-1)-(cosx-1)=0
(cosx-1)(sinx-1)=0
..
cosx=1
x=0
or
sinx=1
x=π/2
..
sin(theta)+cos(theta)=1
√(1-cos^2(x)=1-cosx
1-cos^2(x)=1-2cosx+cos^2(x)
2cos^2x-2cosx=0
2cosx(cosx-1)=0
..
2cosx=0
x=π/2, 2π/3
or
cosx-1=0
cosx=1
x=0
theta is greater than or equal to 0 and less than 2 pi