Solve for x on the interval [0, 2pi]: 4cos(2x)sin(2x)=1

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We have to find the values of x in the interval [0, 2pi] that satisfy 4*cos(2x)*sin(2x) = 1

Use the formula for sin 2x which gives sin 2x = 2*sin x*cos x

Here we have 4*cos(2x)*sin(2x) = 1

=> 2*cos(2x)*sin(2x) = 1/2

=> sin 4x = 1/2

=> 4x = arc sin (1/2)

=> 4x = 30 degrees and 150 degrees

=> x = (1/4)*30 and (1/4)*150

=> x = 7.5 degrees and x = 37.5 degrees

The required solution is x = 7.5 degrees and x = 37.5 degrees

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4*cos(2x)*sin(2x) = 1

First we will divide by 2 .

==> 2cos(2x)*sin(2x) = 1/2................(1)

Now we will use trigonometric identities to solve.

We know that: sin2x = 2sinx*cosx

==> sin4x = 2sin(2x)*cos(2x)

Then we will substitute into (1).

==> sin(4x) = 1/2

But we know that if sin(a)= 1/2 ==> a = pi/6 , and  5pi/6.

==> 4x = pi/6 ==> x = pi/24

==> 4x = 5pi/6 ==> x = 5pi/24

Then the answer is:  x= { pi/24, 5pi/24}

 

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