Question

How do you find all values of x in the interval [0, 2pi] in the equation `2sin^2x - sinx - 1 = 0`?

Answer 1

Solution set in the interval `[0,2pi]` is `{pi/2,(7pi)/6,(11pi)/6}`

Explanation:

To find all values of x in the interval `[0, 2pi]` in the equation `2sin^2x−sinx−1=0`, let us solve by considering `sinx` as variable and using quadratic formula.

Hence `sinx=(-(-1)+-sqrt((-1)^2-4xx2xx(-1)))/(2xx2)` or

`sinx=(1+-sqrt(1+8))/4` i.e.

`sinx=(1+-3)/4` i.e. `sinx=1` or `-1/2`

In the interval `[0,2pi]`, `sinx=1` only for `x=pi/2`

and `sinx=-1/2` for `x=(7pi)/6` and `(11pi)/6`

Hence solution set in the interval `[0,2pi]` is `{pi/2,(7pi)/6,(11pi)/6}`