SOLUTION: Please help me solve this equation. find all solution [0,2pi) 4sin(2x)+4cos(x)=0

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Question 687763: Please help me solve this equation.
find all solution [0,2pi)
4sin(2x)+4cos(x)=0

Answer by fcabanski(1391) About Me  (Show Source):
You can put this solution on YOUR website!
Remember the double angle identity: sin(2a) = 2*sin(a)*cos(a).


4sin(2x)+4cos(x)=0 --->divide out the like term (4) ---> 4(sin(2x)+cos(x)) = 0 ---> (divide both sides by 4)---> sin(2x) + cos(x) = 0 (use the double angle identity.


2sin(x)cos(x) + cos(x) = 0 ---> factor out like terms (cos(x) ---> cos(x)(2sin(x)+1) = 0 Now use the zero product rule.


cos(x) = 0 which for the stated range is 90 degrees or pi/2.


2sin(x) + 1 = 0 ---> 2sin(x) = -1 ---> sin(x) = -1/2 which is 210 degrees or 7pi/6 in the stated range.

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