`sinxcosx = sqrt(3)/4`

To solve, apply the double angle identity `sin(2theta) = 2sinthetacostheta` .

`(sin (2x))/2 =sqrt(3)/4`

Then, multiply both sides by 2 to have sin(2x) only at the left side.

`2*(sin(2x))/2=sqrt3/4*2`

`sin(2x)=sqrt3/2`

In the Unit Circle Chart, sine function has a value of sqrt3/2 at angles pi/3 and 2pi/3....

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`sinxcosx = sqrt(3)/4`

To solve, apply the double angle identity `sin(2theta) = 2sinthetacostheta` .

`(sin (2x))/2 =sqrt(3)/4`

Then, multiply both sides by 2 to have sin(2x) only at the left side.

`2*(sin(2x))/2=sqrt3/4*2`

`sin(2x)=sqrt3/2`

In the Unit Circle Chart, sine function has a value of sqrt3/2 at angles pi/3 and 2pi/3. Since it is a double angle, add each of these two angles by 2pi. So the values of angle 2x are:

`2x = pi/3 , (2pi)/3, (7pi)/3, (8pi)/3`

Then, isolate the x.

`x = pi/6, pi/3, (7pi)/6, (4pi)/3`

Moreover, there is no given interval for x. So the solution is in general form. To express this in general form, notice that the difference between pi/6 and 7pi/6 is pi. And the difference between pi/3 and 4pi/3 is pi.

**Thus, the solution are:**

**`x_1=pi/6+pik` **

**`x_2=pi/3+pik` **