# Find all solutions in the interval [0,2pi) of `4sin^2 x=1`

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### 1 Answer

The equation `4sin^2 x=1` has to be solved for x in [0, 2*pi)

`4*sin^2x = 1`

=> `sin^2x = 1/4`

=> sin x = 1/2 and sin x = -1/2

=> `x = sin^-1(1/2)` and `x = sin^-1(-1/2)`

=> x = 30, x = 150, x = 330 and x = 210

**The required solutions of the equation are 30, 150, 210 and 330 degrees.**