SOLUTION: Solve the following for 0 < x < 2pi
(sin x + cos x)^2 = 1
Algebra.Com
Question 937255: Solve the following for 0 < x < 2pi
(sin x + cos x)^2 = 1
Found 2 solutions by lwsshak3, Alan3354:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Solve the following for 0 < x < 2pi
(sin x + cos x)^2 = 1
(sinx+cosx)^2=sin^2(x)+2sinxcosx+cos^2(x)=1+sin(2x)
1+sin(2x)=1
sin2x=0
2x=0
x=0
no solution: x not in given domain
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Solve the following for 0 < x < 2pi
(sin x + cos x)^2 = 1
sin(x) + cos(x) = 1
By inspection, x = pi/2
------
sin(x) + cos(x) = -1
x = pi
RELATED QUESTIONS
1.Solve the equation 2cos x-1=0, 0 (answered by stanbon)
Please help me solve this equations:
1. {2 sin x-1=0}
2. graph {f(x)=sin(x-2pi)-1}... (answered by stanbon)
Solve the equations of [0, 2pi]:
cosx(2sinx-1)=0
and... (answered by lwsshak3)
Solve on the interval [0, 2pi]
Sin*2x - sin x + 1=... (answered by lwsshak3)
Solve the following equation on the interval [0, 2pi]
cos^2 x + 2 cos x + 1 =... (answered by josgarithmetic)
Hello,Can you please help me solve this problem:
1.Solve the following equation for x... (answered by Edwin McCravy)
Solve the equation for x, on the interval 0 is less than or equal to x and 2pi is greater (answered by Alan3354)
0 2pi: sin^2(x)-sq root of 3... (answered by lwsshak3)
I need to solve the following for x where degrees 0 (answered by Edwin McCravy)