SOLUTION: Solve {{{sin(2x)-cos(x)=0}}} given that 0 &#8804; x < 2&#960;

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Question 1105276: Solve sin%282x%29-cos%28x%29=0 given that 0 ≤ x < 2π
Found 2 solutions by KMST, rothauserc:
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
sin%282x%29=2sin%28x%29cos%28x%29 is a popular trigonometric identity.
sin%282x%29-cos%28x%29=0
2sin%28x%29cos%28x%29-cos%28x%29=0
%282sin%28x%29-1%29cos%28x%29=0
system%282sin%28x%29-1=0%2C%22or%22%2Ccos%28x%29=0%29 --> system%28sin%28x%29=1%2F2%2C%22or%22%2Cx=pi%2F2%2C%22or%22%2Cx=3pi%2F2%29 -->

Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
sin(2x) - cos(x) = 0
:
sin(2x) = cos(x)
:
2sin(x)cos(x) = cos(x)
:
2sin(x) = 1
:
sin(x) = 1/2
:
sin^-1 (1/2) = 30 degrees
:
x = 30 degrees, 150 degrees
:
Note that the sin function is positive in the first and second quadrants
: