SOLUTION: Solve for all values of x in the interval [0, 2𝜋] that satisfy the equation. 4 sin(2x) = 4 cos(x)

Algebra ->  Trigonometry-basics -> SOLUTION: Solve for all values of x in the interval [0, 2𝜋] that satisfy the equation. 4 sin(2x) = 4 cos(x)       Log On


   



Question 1196623: Solve for all values of x in the interval [0, 2𝜋] that satisfy the equation.
4 sin(2x) = 4 cos(x)

Answer by ikleyn(52805) About Me  (Show Source):
You can put this solution on YOUR website!
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Solve for all values of x in the interval [0, 2𝜋] that satisfy the equation.
4 sin(2x) = 4 cos(x)
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4 sin(2x) = 4 cos(x)  implies, by canceling common factor "4" in both sides

    sin(2x) = cos(x),

    2sin(x)*cos(x) = cos(x)

    2sin(x)*cos(x) - cos(x) = 0

    cos(x)*(2sin(x) - 1) = 0


So, EITHER  cos(x) = 0,  OR  sin(x) = 1/2.


If cos(x) = 0,  then  x= pi%2F2  or  x= %283pi%29%2F2.


If sin(x) = 1/2,  then  x= pi%2F6  or  x= pi - pi%2F6 = %285pi%29%2F6.


So, the solution to the given equation are  pi%2F2,  %283pi%29%2F2,  pi%2F6,  %285pi%29%2F6.    ANSWER

Solved.