SOLUTION: Please help me solve this equation: {{{ cos^2(x)+2sin(x)cos(x)-sin^2(x)=0 }}} with the range of x being between 0 and 2pi.

Algebra ->  Trigonometry-basics -> SOLUTION: Please help me solve this equation: {{{ cos^2(x)+2sin(x)cos(x)-sin^2(x)=0 }}} with the range of x being between 0 and 2pi.       Log On


   

Question 1075991: Please help me solve this equation: +cos%5E2%28x%29%2B2sin%28x%29cos%28x%29-sin%5E2%28x%29=0+ with the range of x being between 0 and 2pi.
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
+cos%5E2%28x%29%2B2sin%28x%29cos%28x%29-sin%5E2%28x%29=0+

+cos%5E2%28x%29-sin%5E2%28x%29%2B2sin%28x%29cos%28x%29=0+

Use the formulas for twice an angle:

      cos%282theta%29=cos%5E2%28theta%29-sin%5E2%28theta%29
      sin%282theta%29=2sin%28theta%29cos%28theta%29

cos%282x%29%2Bsin%282x%29=0

cos%282x%29=-sin%282x%29

Divide both sides by cos(2x)

1=-sin%282x%29%2Fcos%282x%29

1=-tan%282x%29

tan%282x%29+=+-1

x is between 0 and 2p, but
2x is between 0 and 4p

so 

and taking half of all those:



Edwin