Question

How do you solve `2cos^2x+sinx+1=0` over the interval 0 to 2pi?

Answer 1

Possible solutions within the domain `[0, 2pi]` are
`x=(3pi)/2`

Explanation:

`2cos^2x+sinx+1=0` can be written as

`2(1-sin^2x)+sinx+1=0`, which can be simplified to

`2sin^2x-sinx-3=0`

As this is an equation of the type `az^2+bz+c=0`, where `z=sinx`, `a=2`, `b=-1` and `c=-3`, whose solution is given by `(-b+-sqrt(b^2-4ac))/(2a)`, we have

`sinx=(-(-1)+-sqrt((-1)^2-4*2*(-3)))/(2.2)` or

`sinx=(1+-sqrt25)/4` or

`sinx=(1+-5)/4` i.e. `sinx= 3/2 or -1`

As range of `sinx` is within [-1,1}, it cannot be `3/2` and `-1` isonly possibility.

Hence, possible solution within the domain `[0, 2pi]` are

`x=(3pi)/2`