Question

Can someone solve this, please?

`3sin ^(2) (2x) = 1`
when x€[0,2pi]​​

Answer 1

9514 1404 393

Answer:

x ∈ {{0.30773985, 1.26305647, 1.87853618, 2.83385280, 3.44933251, 4.40464913, 5.02012883, 5.97544545}

Explanation:

Divide by the coefficient of the sine function, take the root, then use the arcsine function to find the angles.

`3\sin^2(2x)=1\\\\\sin^2(2x)=(1)/(3)\\\\ \sin(2x)=\pm\sqrt{(1)/(3)}\\\\2x=\pm\arcsin\left((√(3))/(3)\right)+n\pi\\\\x=(\pm\arcsin\left((√(3))/(3)\right)+n\pi)/(2)`

The numerical values are approximately ...

x ∈ {{0.30773985, 1.26305647, 1.87853618, 2.83385280, 3.44933251, 4.40464913, 5.02012883, 5.97544545}

__

The graph shows the zeros of 3sin²(2x) -1.