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# TRIG EQUATION!!! NEED HELP?

solve for x between 0 and 2pie, algebraiclly

(sinx+cosx)^2=0

PLZ HELP

### 1 Answer

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- 9 years agoFavorite Answer
There are 2 ways to solve this:

(sin(x) + cos(x))^2 = 0

sin(x) + cos(x) = sqrt(0)

sin(x) + cos(x) = 0

sin(x) = -cos(x)

sin(x)/cos(x) = -1

tan(x) = -1

x = 3pi/4 , 7pi/4

(sin(x) + cos(x))^2 = 0

sin(x)^2 + 2sin(x)cos(x) + cos(x)^2 = 0

(sin(x)^2 + cos(x)^2) + 2 * sin(x) * cos(x) = 0

1 + sin(2x) = 0

sin(2x) = -1

2x = 3pi/2 , 3pi/2 + 2pi

2x = 3pi/2 , 7pi/2

x = 3pi/4 , 7pi/4

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