Solve the equation, X is real number. 2cos(6x) + 4sin(3x) = 3

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hala718 | High School Teacher | (Level 1) Educator Emeritus

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2cos(6x) + 3sin(3x)= 3

First, we know that:

cos2x = 2sin^2 x - 1

==> cos6x = 1-2sin^2 3x

Now we will substitute into the equation.

==> 2(1-2sin^2 3x) + 4sin3x = 3

==> 2 - 4sin^2 3x + 4sin3x -3 = 0

==> -4sin62 3x + 4sin3x -1 = 0

Multiply by -1:

==> 4sin^2 3x - 4sin3x +1 = 0

Now we will factor:

==> (2sin3x -1)^2 = 0

==> (2sin3x -1)= 0

==> sin3x = 1/2

==> 3x = pi/6+2npi , 5pi/6+2npi  (n= 0, 1, 2, 3...)

==> x = pi/18 , 5pi/18

==> x = { pi/18+2npi , 5pi/18+2npi}

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